RS Aggarwal Solutions for Class 11 Chapter 04 Principle Of Mathematical Induction

Question 1. Using the principle of mathematical induction, prove each of the following for all n ∈ N: 1 + 2 + 3 + 4 + … + n = 1/2 n(n + 1)
Answer: We need to show that \( 1 + 2 + 3 + 4 + … + n = \frac{1}{2}n(n + 1) \).

Steps to prove by mathematical induction:

Define P(n) as a statement involving the natural number n such that
(i) P(1) is true
(ii) P(k + 1) is true, whenever P(k) is true

Then P(n) is true for all n ∈ N

Let P(n): \( 1 + 2 + 3 + 4 + … + n = \frac{1}{2}n(n + 1) \)

Step 1:

\( P(1) = \frac{1}{2} \times 1 \times (1 + 1) = \frac{1}{2} \times 2 = 1 \)

Therefore, P(1) is true

Step 2:

Assume P(k) is true. Then,

\( P(k): 1 + 2 + 3 + 4 + … + k = \frac{1}{2}k(k + 1) \)

Now,

\( 1 + 2 + 3 + 4 + … + k + (k + 1) = \frac{1}{2}k(k + 1) + (k + 1) \)

\( = (k + 1)\left\{\frac{1}{2}k + 1\right\} \)

\( = \frac{1}{2}(k + 1)(k + 2) \)

\( = P(k + 1) \)

Hence, P(k + 1) is true whenever P(k) is true

By the principle of mathematical induction, \( 1 + 2 + 3 + 4 + … + n = \frac{1}{2}n(n + 1) \) for all n ∈ N

Hence proved.

Exam Tip: Always check the base case carefully - it must be verified before assuming the inductive step. The inductive hypothesis assumes P(k) is true and must be used to prove P(k + 1).

Question 2. Using the principle of mathematical induction, prove each of the following for all n ∈ N: 2 + 4 + 6 + 8 + … + 2n = n(n + 1)
Answer: We need to show that \( 2 + 4 + 6 + 8 + … + 2n = n(n + 1) \).

Steps to prove by mathematical induction:

Define P(n) as a statement involving the natural number n such that
(i) P(1) is true
(ii) P(k + 1) is true, whenever P(k) is true

Then P(n) is true for all n ∈ N

Let P(n): \( 2 + 4 + 6 + 8 + … + 2n = n(n + 1) \)

Step 1:

\( P(1) = 1(1 + 1) = 1 \times 2 = 2 \)

Therefore, P(1) is true

Step 2:

Assume P(k) is true. Then,

\( P(k): 2 + 4 + 6 + 8 + … + 2k = k(k + 1) \)

Now,

\( 2 + 4 + 6 + 8 + … + 2k + 2(k + 1) = k(k + 1) + 2(k + 1) \)

\( = (k + 1)(k + 2) \)

\( = P(k + 1) \)

Hence, P(k + 1) is true whenever P(k) is true

By the principle of mathematical induction, \( 2 + 4 + 6 + 8 + … + 2n = n(n + 1) \) for all n ∈ N

Exam Tip: Notice that this is the sum of the first n even numbers - factoring out 2 from each term shows it equals 2 times the sum of first n natural numbers.

Question 3. Using the principle of mathematical induction, prove each of the following for all n ∈ N: \( 1 + 3 + 3^2 + 3^3 + … + 3^{n-1} = \frac{1}{2}(3^n - 1) \)
Answer: We need to show that \( 1 + 3 + 3^2 + 3^3 + … + 3^{n-1} = \frac{3^n - 1}{2} \).

Steps to prove by mathematical induction:

Define P(n) as a statement involving the natural number n such that
(i) P(1) is true
(ii) P(k + 1) is true, whenever P(k) is true

Then P(n) is true for all n ∈ N

Let P(n): \( 1 + 3 + 3^2 + … + 3^{n-1} = \frac{3^n - 1}{2} \)

Step 1:

\( P(1) = \frac{3^1 - 1}{2} = \frac{2}{2} = 1 \)

Therefore, P(1) is true

Step 2:

Assume P(k) is true. Then,

\( P(k): 1 + 3 + 3^2 + … + 3^{k-1} = \frac{3^k - 1}{2} \)

Now,

\( 1 + 3 + 3^2 + … + 3^{k-1} + 3^{(k+1)-1} = \frac{3^k - 1}{2} + 3^k \)

\( = \frac{3^k - 1}{2} + 3^k \)

\( = 3^k\left(\frac{1}{2} + 1\right) - \frac{1}{2} \)

\( = 3^k\left(\frac{3}{2}\right) - \frac{1}{2} \)

\( = 3^{k+1}\left(\frac{3}{2}\right) - \frac{1}{2} \)

\( = \frac{3^{k+1} - 1}{2} \)

\( = P(k + 1) \)

Hence, P(k + 1) is true whenever P(k) is true

By the principle of mathematical induction, \( 1 + 3 + 3^2 + … + 3^{n-1} = \frac{3^n - 1}{2} \) for all n ∈ N

Exam Tip: This is a geometric series with first term 1 and common ratio 3 - recognize the formula pattern to simplify calculations.

Question 4. Using the principle of mathematical induction, prove each of the following for all n ∈ N: \( 2 + 6 + 18 + … + 2 \times 3^{n-1} = (3^n - 1) \)
Answer: We need to show that \( 2 + 6 + 18 + … + 2 \times 3^{n-1} = 3^n - 1 \).

Steps to prove by mathematical induction:

Define P(n) as a statement involving the natural number n such that
(i) P(1) is true
(ii) P(k + 1) is true, whenever P(k) is true

Then P(n) is true for all n ∈ N

Let P(n): \( 2 + 6 + 18 + … + 2 \times 3^{n-1} = 3^n - 1 \)

Step 1:

\( P(1) = 3^1 - 1 = 3 - 1 = 2 \)

Therefore, P(1) is true

Step 2:

Assume P(k) is true. Then,

\( P(k): 2 + 6 + 18 + … + 2 \times 3^{k-1} = 3^k - 1 \)

Now,

\( 2 + 6 + 18 + … + 2 \times 3^{k-1} + 2 \times 3^{k+1-1} = 3^k - 1 + 2 \times 3^k \)

\( = - 1 + 3 \times 3^k \)

\( = 3^{k+1} - 1 \)

\( = P(k + 1) \)

Hence, P(k + 1) is true whenever P(k) is true

By the principle of mathematical induction, \( 2 + 6 + 18 + … + 2 \times 3^{n-1} = 3^n - 1 \) for all n ∈ N

Exam Tip: Factor out the common pattern - each term is 2 times a power of 3, forming a geometric series that simplifies elegantly.

Question 5. Using the principle of mathematical induction, prove each of the following for all n ∈ N: \( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + … + \frac{1}{2^n} = \left(1 - \frac{1}{2^n}\right) \)
Answer: We need to show that \( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + … + \frac{1}{2^n} = 1 - \frac{1}{2^n} \).

Steps to prove by mathematical induction:

Define P(n) as a statement involving the natural number n such that
(i) P(1) is true
(ii) P(k + 1) is true, whenever P(k) is true

Then P(n) is true for all n ∈ N

Let P(n): \( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + … + \frac{1}{2^n} = 1 - \frac{1}{2^n} \)

Step 1:

\( P(1) = 1 - \frac{1}{2^1} = 1 - \frac{1}{2} = \frac{1}{2} \)

Therefore, P(1) is true

Step 2:

Assume P(k) is true. Then,

\( P(k): \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + … + \frac{1}{2^k} = 1 - \frac{1}{2^k} \)

Now,

\( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + … + \frac{1}{2^k} + \frac{1}{2^{k+1}} = 1 - \frac{1}{2^k} + \frac{1}{2^{k+1}} \)

\( = 1 - \frac{1}{2^k} + \frac{1}{2^{k+1}} \)

\( = 1 + \frac{1}{2^k}\left(\frac{1}{2} - 1\right) \)

\( = 1 + \frac{1}{2^k}\left(-\frac{1}{2}\right) \)

\( = 1 - \frac{1}{2^{k+1}} \)

\( = P(k + 1) \)

Hence, P(k + 1) is true whenever P(k) is true

By the principle of mathematical induction, \( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + … + \frac{1}{2^n} = 1 - \frac{1}{2^n} \) for all n ∈ N

Exam Tip: This is a geometric series with first term 1/2 and common ratio 1/2 - as n increases, the sum approaches 1 from below.

Question 6. Using the principle of mathematical induction, prove each of the following for all n ∈ N: \( 1^2 + 3^2 + 5^2 + 7^2 + … + (2n - 1)^2 = \frac{n(2n - 1)(2n + 1)}{3} \)
Answer: We need to show that \( 1^2 + 3^2 + 5^2 + 7^2 + … + (2n - 1)^2 = \frac{n(2n - 1)(2n + 1)}{3} \).

Steps to prove by mathematical induction:

Define P(n) as a statement involving the natural number n such that
(i) P(1) is true
(ii) P(k + 1) is true, whenever P(k) is true

Then P(n) is true for all n ∈ N

Let P(n): \( 1^2 + 3^2 + 5^2 + 7^2 + … + (2n - 1)^2 = \frac{n(2n - 1)(2n + 1)}{3} \)

Step 1:

\( P(1) = \frac{1(2 - 1)(2 + 1)}{3} = \frac{3}{3} = 1 \)

Therefore, P(1) is true

Step 2:

Assume P(k) is true. Then,

\( P(k): 1^2 + 3^2 + 5^2 + 7^2 + … + (2k - 1)^2 = \frac{k(2k - 1)(2k + 1)}{3} \)

Now,

\( 1^2 + 3^2 + 5^2 + 7^2 + … + (2(k + 1) - 1)^2 = \frac{k(2k - 1)(2k + 1)}{3} + (2k + 1)^2 \)

\( = \frac{k(2k - 1)(2k + 1)}{3} + (2k + 1)^2 \)

\( = (2k + 1)\left[\frac{k(2k - 1)}{3} + (2k + 1)\right] \)

\( = (2k + 1)\left[\frac{2k^2 - k + 6k + 3}{3}\right] \)

\( = (2k + 1)\left[\frac{2k^2 + 5k + 3}{3}\right] \)

\( = (2k + 1)\left[\frac{(k + 1)(2k + 3)}{3}\right] \) (Splitting the middle term)

\( = \frac{(k + 1)(2k + 1)(2k + 3)}{3} \)

\( = P(k + 1) \)

Hence, P(k + 1) is true whenever P(k) is true

By the principle of mathematical induction, \( 1^2 + 3^2 + 5^2 + 7^2 + … + (2n - 1)^2 = \frac{n(2n - 1)(2n + 1)}{3} \) for all n ∈ N

Exam Tip: The sum of squares of odd numbers follows a specific pattern - factor expressions carefully and use algebraic identities to simplify.

Question 7. Using the principle of mathematical induction, prove each of the following for all n ∈ N: \( 1 \times 2^1 + 2 \times 2^2 + 3 \times 2^3 + … + n \times 2^n = (n - 1) \times 2^{n+1} + 2 \)
Answer: We need to show that \( 1 \times 2^1 + 2 \times 2^2 + 3 \times 2^3 + … + n \times 2^n = (n - 1) \times 2^{n+1} + 2 \).

Proof by principle of mathematical induction (PMI):

Let P(n): \( 1 \times 2^1 + 2 \times 2^2 + 3 \times 2^3 + … + n \times 2^n \)

For n = 1

LHS = \( 1 \times 2 = 2 \)

RHS = \( (1 - 1) \times 2^{(1+1)} + 2 = 0 + 2 = 2 \)

Hence, LHS = RHS

P(n) is true for n = 1

Assume P(k) is true

\( 1 \times 2^1 + 2 \times 2^2 + 3 \times 2^3 + k \times 2^k = (k - 1) \times 2^{k+1} + 2 \) ... (1)

We will prove that P(k + 1) is true

\( 1 \times 2^1 + 2 \times 2^2 + 3 \times 2^3 + (k + 1) \times 2^{k+1} = ((k + 1) - 1) \times 2^{(k+1)+1} + 2 = k \times 2^{k+2} + 2 \) ... (2)

From (1)

\( 1 \times 2^1 + 2 \times 2^2 + 3 \times 2^3 + k \times 2^k = (k - 1) \times 2^{k+1} + 2 \)

Adding \( (k + 1) \times 2^{k+1} \) to both sides,

\( (1 \times 2^1 + 2 \times 2^2 + 3 \times 2^3 + k \times 2^k) + (k + 1) \times 2^{k+1} = (k - 1) \times 2^{k+1} + 2 + (k + 1) \times 2^{k+1} \)

\( = k \times 2^{k+1} - 2^{k+1} + 2 + k \times 2^{k+1} + 2^{k+1} \)

\( = 2k \times 2^{k+1} + 2 \)

\( = k \times 2^{k+2} + 2 \)

\( (1 \times 2^1 + 2 \times 2^2 + 3 \times 2^3 + k \times 2^k) + (k + 1) \times 2^{k+1} = k \times 2^{k+2} + 2 \)

Which is the same as P(k + 1)

Therefore, P(k + 1) is true whenever P(k) is true

By the principle of mathematical induction, P(n) is true for all natural numbers n

Put k = n - 1

\( 1 \times 2^1 + 2 \times 2^2 + 3 \times 2^3 + n \times 2^n = (n - 1) \times 2^{n+1} + 2 \)

Hence proved.

Exam Tip: When coefficients increase linearly (1, 2, 3, ..., n) with exponential terms, separate the inductive hypothesis sum and the new term, then combine algebraically.

Question 8. Using the principle of mathematical induction, prove each of the following for all n ∈ N: \( 3 \times 2^2 + 3^2 \times 2^3 + 3^3 \times 2^4 + … + 3^n \times 2^{n+1} = \frac{12}{5}(6^n - 1) \)
Answer: We need to show that \( 3 \times 2^2 + 3^2 \times 2^3 + 3^3 \times 2^4 + … + 3^n \times 2^{n+1} = \frac{12}{5}(6^n - 1) \).

Proof by principle of mathematical induction (PMI):

Let P(n): \( 3 \times 2^2 + 3^2 \times 2^3 + 3^3 \times 2^4 + … + 3^n \times 2^{n+1} \)

For n = 1

LHS = \( 3 \times 2^2 = 12 \)

RHS = \( \frac{12}{5} \times (6^1 - 1) = \frac{12}{5} \times 5 = 12 \)

Hence, LHS = RHS

P(n) is true for n = 1

Assume P(k) is true

\( 3 \times 2^2 + 3^2 \times 2^3 + 3^3 \times 2^4 + … + 3^k \times 2^{k+1} = \frac{12}{5}(6^k - 1) \) ... (1)

We will prove that P(k + 1) is true

\( 3 \times 2^2 + 3^2 \times 2^3 + 3^3 \times 2^4 + … + 3^{k+1} \times 2^{k+2} = \frac{12}{5}(6^{k+1} - 1) \)

\( = \frac{12}{5}(6^{k+1}) - \frac{12}{5} \) ... (2)

We have to prove P(k + 1) from P(k), i.e. (2) from (1)

From (1)

\( 3 \times 2^2 + 3^2 \times 2^3 + 3^3 \times 2^4 + … + 3^k \times 2^{k+1} = \frac{12}{5}(6^k - 1) \)

Adding \( 3^{k+1} \times 2^{k+2} \) to both sides,

\( 3 \times 2^2 + 3^2 \times 2^3 + 3^3 \times 2^4 + … + 3^k \times 2^{k+1} + 3^{k+1} \times 2^{k+2} = \frac{12}{5}(6^k - 1) + 3^{k+1} \times 2^{k+2} \)

\( = \frac{12}{5}(6^k - 1) + 3^k \times 2^k \times 12 \)

\( = \frac{12}{5}(6^k - 1) + 6^k \times 12 \)

\( = 6^k\left(\frac{12}{5} + 12\right) - \frac{12}{5} \)

\( = \left(\frac{72}{5}\right) - \frac{12}{5} \)

\( = \frac{12}{5}(6^{k+1}) - \frac{12}{5} \)

\( 3 \times 2^2 + 3^2 \times 2^3 + 3^3 \times 2^4 + … + 3^k \times 2^{k+1} + 3^{k+1} \times 2^{k+2} = \frac{12}{5}(6^{k+1}) - \frac{12}{5} \)

Which is the same as P(k + 1)

Therefore, P(k + 1) is true whenever P(k) is true.

By the principle of mathematical induction, P(n) is true for all natural numbers n

Put k = n - 1

\( 3 \times 2^2 + 3^2 \times 2^3 + 3^3 \times 2^4 + … + 3^n \times 2^{n+1} = \frac{12}{5}(6^n - 1) \)

Hence proved

Exam Tip: Recognize that \( 3 \times 2 = 6 \), so each term is a power of 6 multiplied by a constant - this simplifies the algebra significantly in the inductive step.

Question 9. Using the principle of mathematical induction, prove each of the following for all n ∈ N: \( 1 + \frac{1}{(1+2)} + \frac{1}{(1+2+3)} + … + \frac{1}{(1+2+3+…+n)} = \frac{2n}{(n+1)} \)
Answer: We need to show that \( 1 + \frac{1}{(1+2)} + \frac{1}{(1+2+3)} + … + \frac{1}{(1+2+3+…+n)} = \frac{2n}{(n+1)} \).

Proof by principle of mathematical induction (PMI):

Let P(n): \( 1 + \frac{1}{(1+2)} + \frac{1}{(1+2+3)} + … + \frac{1}{(1+2+3+…+n)} = \frac{2n}{(n+1)} \)

For n = 1

LHS = 1

RHS = \( \frac{2 \times 1}{(1 + 1)} = 1 \)

Hence, LHS = RHS

P(n) is true for n = 1

Assume P(k) is true

\( 1 + \frac{1}{(1+2)} + … + \frac{1}{(1+2+3+…+k)} = \frac{2k}{(k+1)} \) ... (1)

We will prove that P(k + 1) is true

RHS = \( \frac{2(k+1)}{(k + 1 + 1)} = \frac{2k + 2}{k + 2} \)

LHS =

\( 1 + \frac{1}{(1+2)} + … + \frac{1}{(1+2+3+…+(k+1))} \)

\( = 1 + \frac{1}{(1+2)} + … + \frac{1}{(1+2+3+…+k)} + \frac{1}{(1+2+3+…+(k+1))} \) [Writing the last second term]

\( = \frac{2k}{(k + 1)} + \frac{1}{(1+2+3+…+(k+1))} \) [From (1)]

\( = \frac{2k}{(k + 1)} + \frac{1}{(k + 1) \times (k + 2) / 2} \) [Since \( 1 + 2 + 3 + … + n = \frac{n(n+1)}{2} \), putting n = k + 1]

\( = \frac{2k}{(k + 1)} + \frac{2}{(k + 1) \times (k + 2)} \)

\( = \frac{2}{(k + 1)} \left(\frac{k}{1} + \frac{1}{k + 2}\right) \)

\( = \frac{2}{k + 1} \left(\frac{k(k + 1) \times (k + 1)}{k + 2}\right) \)

\( = \frac{2(k + 1)}{k + 2} \) [Taking LCM and simplifying]

\( = \frac{2(k + 1)}{(k + 2)} \)

\( = \) RHS

Therefore, LHS = RHS

Therefore, P(k + 1) is true whenever P(k) is true.

By the principle of mathematical induction, P(n) is true for all natural numbers n

Exam Tip: Recognize that the sum of first m natural numbers equals m(m+1)/2 - use this to convert each denominator into a simple fraction before combining terms.

Question 10. Using the principle of mathematical induction, prove each of the following for all n ϵ N:
\( \frac{1}{2 \times 5} + \frac{1}{(5 \times 8)} + \ldots + \frac{1}{(3n-1) \times (3n+2)} = \frac{n}{(6n+4)} \)
Answer: We need to establish the statement using PMI.

For the base case where n = 1:

LHS = \( \frac{1}{2 \times 5} = \frac{1}{10} \)

RHS = \( \frac{1}{(6+4)} = \frac{1}{10} \)

Since LHS equals RHS, P(1) holds true.

Now we assume P(k) is valid for some positive integer k:
\( \frac{1}{2 \times 5} + \frac{1}{(5 \times 8)} + \ldots + \frac{1}{(3k-1) \times (3k+2)} = \frac{k}{(6k+4)} \) ...(1)

To show P(k + 1) is valid, we need to demonstrate:

RHS = \( \frac{k+1}{(6(k+1)+4)} = \frac{k+1}{(6k+10)} \)

LHS = \( \frac{1}{2 \times 5} + \frac{1}{(5 \times 8)} + \ldots + \frac{1}{(3k-1) \times (3k+2)} + \frac{1}{(3(k+1)-1) \times (3(k+1)+2)} \)

Rearranging the last term in the sum:

= \( \frac{1}{2 \times 5} + \frac{1}{(5 \times 8)} + \ldots + \frac{1}{(3k-1) \times (3k+2)} + \frac{1}{(3k+2) \times (3k+5)} \)

Using equation (1):

= \( \frac{k}{(6k+4)} + \frac{1}{(3k+2) \times (3k+5)} \)

= \( \frac{k}{(6k+4)} + \frac{1}{(3k+2)(3k+5)} \)

Factoring out common terms and simplifying the numerator:

= \( \frac{1}{(3k+2)} \times \left[\frac{(3k+5)(k+1)}{2(3k+5)} \right] \) (Taking LCM and simplifying)

= \( \frac{k+1}{(6k+10)} \)

= RHS

Therefore, LHS = RHS, proving P(k + 1) is valid when P(k) holds.

By mathematical induction, P(n) is valid for every natural number n.

Setting k = n - 1 confirms the final result:
\( \frac{1}{2 \times 5} + \frac{1}{(5 \times 8)} + \ldots + \frac{1}{(3n-1) \times (3n+2)} = \frac{n}{(6n+4)} \)

Hence proved.
In simple words: The sum of fractions with denominators following the pattern \( (3r-1)(3r+2) \) equals \( \frac{n}{6n+4} \) when you add them all up to the nth term. We confirm this is correct by checking the first case, assuming it works for k terms, and showing it must work for k + 1 terms as well.

Exam Tip: Always combine fractions by finding a common denominator before applying the induction hypothesis. Verify algebraic cancellations step by step to avoid careless errors.

Question 11. Using the principle of mathematical induction, prove each of the following for all n ϵ N:
\( \frac{1}{1.4} + \frac{1}{4.7} + \frac{1}{7.10} + \ldots + \frac{1}{(3n-2)(3n+1)} = \frac{n}{(3n+1)} \)
Answer: We establish this statement using PMI.

For n = 1 (base case):

LHS = \( \frac{1}{1 \times 4} = \frac{1}{4} \)

RHS = \( \frac{1}{(3+1)} = \frac{1}{4} \)

Since LHS = RHS, P(1) is valid.

Assume P(k) holds for some positive integer k:
\( \frac{1}{1 \times 4} + \frac{1}{(4 \times 7)} + \ldots + \frac{1}{(3k-2) \times (3k+1)} = \frac{k}{(3k+1)} \) ...(1)

We prove P(k + 1) is valid:

RHS = \( \frac{k+1}{(3(k+1)+1)} = \frac{k+1}{(3k+4)} \)

LHS = \( \frac{1}{1 \times 4} + \frac{1}{(4 \times 7)} + \ldots + \frac{1}{(3k-2) \times (3k+1)} + \frac{1}{(3(k+1)-2) \times (3(k+1)+1)} \)

Expanding the last term:

= \( \frac{1}{1 \times 4} + \frac{1}{(4 \times 7)} + \ldots + \frac{1}{(3k-2) \times (3k+1)} + \frac{1}{(3k+1) \times (3k+4)} \)

Using equation (1):

= \( \frac{k}{(3k+1)} + \frac{1}{(3k+1) \times (3k+4)} \) [Using 1]

= \( \frac{k}{(6k+4)} + \frac{1}{(3k+2)(3k+5)} \)

Factoring out and combining:

= \( \frac{1}{(3k+1)} \left(k + \frac{1}{(3k+4)} \right) \)

= \( \frac{1}{(3k+1)} \times \left[\frac{(3k+5)(k+1)}{2(3k+5)} \right] \) (Splitting and simplifying)

= \( \frac{k+1}{(3k+4)} \)

= RHS

Therefore P(k + 1) is valid whenever P(k) holds.

By mathematical induction, P(n) is valid for all natural numbers.

Hence proved.
In simple words: This sum follows a telescoping pattern where consecutive fractions partially cancel out, leaving you with a simple final form of \( \frac{n}{3n+1} \).

Exam Tip: Recognise telescoping fractions early - they simplify dramatically through cancellation, making induction proofs cleaner and faster.

Question 12. Using the principle of mathematical induction, prove each of the following for all n ϵ N:
\( \frac{1}{1.3} + \frac{1}{3.5} + \frac{1}{5.7} + \ldots + \frac{1}{(2n-1)(2n+1)} = \frac{n}{(2n+1)} \)
Answer: We prove this using PMI.

For n = 1 (base case):

LHS = \( \frac{1}{1 \times 3} = \frac{1}{3} \)

RHS = \( \frac{1}{(2+1)} = \frac{1}{3} \)

So P(1) holds true.

Assume P(k) is valid for some positive integer k:
\( \frac{1}{1 \times 3} + \frac{1}{(3 \times 5)} + \ldots + \frac{1}{(2k-1) \times (2k+1)} = \frac{k}{(2k+1)} \) ...(1)

To verify P(k + 1):

RHS = \( \frac{k+1}{(2(k+1)+1)} = \frac{k+1}{(2k+3)} \)

LHS = \( \frac{1}{1 \times 3} + \frac{1}{(3 \times 5)} + \ldots + \frac{1}{(2k-1) \times (2k+1)} + \frac{1}{(2(k+1)-1) \times (2(k+1)+1)} \)

Rewriting the final term:

= \( \frac{1}{1 \times 3} + \frac{1}{(3 \times 5)} + \ldots + \frac{1}{(2k-1) \times (2k+1)} + \frac{1}{(2k+1) \times (2k+3)} \)

Using (1):

= \( \frac{k}{(2k+1)} + \frac{1}{(2k+1) \times (2k+3)} \) [Using 1]

= \( \frac{1}{(2k+1)} \left(k + \frac{1}{(2k+3)} \right) \)

= \( \frac{1}{(2k+1)} \times \left[\frac{(2k^2+3k+1)}{(2k+3)} \right] \)

Factoring the numerator and cancelling:

= \( \frac{k+1}{(2k+3)} \) (Splitting and cancelling)

= RHS

Therefore P(k + 1) is valid whenever P(k) holds.

By mathematical induction, P(n) is valid for all natural numbers n.

Hence proved.
In simple words: Each fraction has a denominator that is the product of two consecutive odd numbers. When you sum them using the telescope method, most terms cancel, giving you a clean result.

Exam Tip: For products of consecutive integers in denominators, always try partial fraction decomposition - it often reveals a telescoping structure that makes the proof much simpler.

Question 13. Using the principle of mathematical induction, prove each of the following for all n ϵ N:
\( \frac{1}{2 \times 5} + \frac{1}{(5 \times 8)} + \ldots + \frac{1}{(3n-1) \times (3n+2)} = \frac{n}{(6n+4)} \)
Answer: We establish this result using PMI.

For n = 1 (base case):

LHS = \( \frac{1}{2 \times 5} = \frac{1}{10} \)

RHS = \( \frac{1}{(6+4)} = \frac{1}{10} \)

Thus P(1) is valid.

Assume P(k) holds for some positive integer k:
\( \frac{1}{2 \times 5} + \frac{1}{(5 \times 8)} + \ldots + \frac{1}{(3k-1) \times (3k+2)} = \frac{k}{(6k+4)} \) ...(1)

To prove P(k + 1):

RHS = \( \frac{k+1}{(6(k+1)+4)} = \frac{k+1}{(6k+10)} \)

LHS = \( \frac{1}{2 \times 5} + \frac{1}{(5 \times 8)} + \ldots + \frac{1}{(3k-1) \times (3k+2)} + \frac{1}{(3(k+1)-1) \times (3(k+1)+2)} \)

Substituting the next term:

= \( \frac{1}{2 \times 5} + \frac{1}{(5 \times 8)} + \ldots + \frac{1}{(3k-1) \times (3k+2)} + \frac{1}{(3k+2) \times (3k+5)} \)

From (1):

= \( \frac{k}{(6k+4)} + \frac{1}{(3k+2) \times (3k+5)} \) [Using 1]

Combining over a common denominator:

= \( \frac{k}{(6k+4)} + \frac{1}{(3k+2)(3k+5)} \)

= \( \frac{1}{(3k+2)} \times \left[\frac{(3k+5)(k+1)}{2(3k+5)} \right] \) (Taking LCM and simplifying)

= \( \frac{k+1}{(6k+10)} \)

= RHS

Therefore P(k + 1) is valid when P(k) holds true.

By mathematical induction, P(n) is valid for all natural numbers n.

Hence proved.
In simple words: The sum telescopes nicely because the denominators follow a pattern where consecutive terms share a common factor, allowing cancellation.

Exam Tip: Always check whether the general term permits partial fraction expansion - this reveals whether telescoping will occur and greatly simplifies your proof.

Question 14. Using the principle of mathematical induction, prove each of the following for all n ϵ N:
\( \left(1 + \frac{3}{1}\right) \times \left(1 + \frac{5}{4}\right) \times \left(1 + \frac{7}{9}\right) \times \ldots \times \left\{1 + \frac{(2n+1)}{n^2}\right\} = (n+1)^2 \)
Answer: We establish this claim using PMI.

For n = 1 (base case):

LHS = \( 1 + \frac{3}{1} = 4 \)

RHS = \( (1+1)^2 = 4 \)

Since LHS = RHS, P(1) is valid.

Assume P(k) holds true for some positive integer k:
\( \left(1 + \frac{3}{1}\right) \times \left(1 + \frac{5}{4}\right) \times \left(1 + \frac{7}{9}\right) \times \ldots \times \left\{1 + \frac{2k+1}{k^2}\right\} = (k+1)^2 \) ...(1)

We now prove P(k + 1):

RHS = \( ((k+1)+1)^2 = (k+2)^2 \)

LHS = \( \left(1 + \frac{3}{1}\right) \times \left(1 + \frac{5}{4}\right) \times \left(1 + \frac{7}{9}\right) \times \ldots \times \left\{1 + \frac{2k+1}{k^2}\right\} \times \left\{1 + \frac{2(k+1)+1}{(k+1)^2}\right\} \)

Rewriting the last factor:

= \( \left(1 + \frac{3}{1}\right) \times \left(1 + \frac{5}{4}\right) \times \left(1 + \frac{7}{9}\right) \times \ldots \times \left\{1 + \frac{2k+1}{k^2}\right\} \times \left\{1 + \frac{2(k+1)+1}{(k+1)^2}\right\} \)

Using equation (1):

= \( (k+1)^2 \times \left\{1 + \frac{2(k+1)+1}{(k+1)^2}\right\} \) [Using 1]

= \( (k+1)^2 \times \left\{1 + \frac{(2k+3)}{(k+1)^2}\right\) \)

= \( (k+1)^2 \times \left\{\frac{(k+1)^2 + (2k+3)}{(k+1)^2}\right\) \)

Expanding the numerator:

= \( (k+1)^2 + (2k+3) \)

= \( k^2 + 2k + 1 + 2k + 3 \)

= \( k^2 + 4k + 4 \)

= \( (k+2)^2 \)

= RHS

Therefore P(k + 1) is valid when P(k) holds.

By mathematical induction, P(n) is valid for all natural numbers n.

Hence proved.
In simple words: Each factor in the product can be rewritten as a fraction. When you multiply the induction hypothesis by the next factor, the algebra simplifies neatly to give you a perfect square.

Exam Tip: When dealing with products, always simplify each factor algebraically before substituting the induction hypothesis - this often reveals hidden cancellations or perfect square patterns.

Question 15. Using the principle of mathematical induction, prove each of the following for all n ϵ N:
\( \left(1 + \frac{1}{1}\right) \times \left(1 + \frac{1}{2}\right) \times \left(1 + \frac{1}{3}\right) \times \ldots \times \left\{1 + \frac{1}{n}\right\} = (n+1) \)
Answer: We prove this statement using PMI.

For n = 1 (base case):

LHS = \( 1 + \frac{1}{1} = 2 \)

RHS = \( (1+1)^1 = 2 \)

Therefore P(1) is valid.

Assume P(k) holds for some positive integer k:
\( \left(1 + \frac{1}{1}\right) \times \left(1 + \frac{1}{2}\right) \times \left(1 + \frac{1}{3}\right) \times \ldots \times \left\{1 + \frac{1}{k}\right\} = (k+1)^1 \) ...(1)

To establish P(k + 1):

RHS = \( ((k+1)+1)^1 = (k+2)^1 = k + 2 \)

LHS = \( \left(1 + \frac{1}{1}\right) \times \left(1 + \frac{1}{2}\right) \times \left(1 + \frac{1}{3}\right) \times \ldots \times \left\{1 + \frac{1}{k}\right\} \times \left\{1 + \frac{1}{(k+1)}\right\} \)

Grouping the first k factors:

= \( \left(1 + \frac{1}{1}\right) \times \left(1 + \frac{1}{2}\right) \times \left(1 + \frac{1}{3}\right) \times \ldots \times \left\{1 + \frac{1}{k}\right\} \times \left\{1 + \frac{1}{(k+1)}\right\} \)

Using (1):

= \( (k+1) \times \left\{1 + \frac{1}{(k+1)}\right\) \) [Using 1]

= \( (k+1) \times \left\{\frac{(k+1)+1}{(k+1)}\right\) \)

= \( (k+1) \times \left\{\frac{(k+2)}{(k+1)}\right\) \)

= \( k + 2 \)

= RHS

Therefore P(k + 1) is valid when P(k) holds true.

By mathematical induction, P(n) is valid for all natural numbers n.

Hence proved.
In simple words: This is a product of simple fractions. When you multiply (k + 1) by the factor (1 + 1/(k + 1)), you get (k + 2) directly because the arithmetic cancels nicely.

Exam Tip: For product proofs, factor out and simplify each term before applying the induction hypothesis to avoid unnecessary algebra and keep the calculation transparent.

Question 16. Using the principle of mathematical induction, prove each of the following for all n ϵ N: n × (n + 1) × (n + 2) is multiple of 6
Answer: We establish this using PMI for all natural numbers.

For n = 1 (base case), we check that P(1) holds:
\( 1 \times (1+1) \times (1+2) = 1 \times 2 \times 3 = 6 \), which is a multiple of 6, so P(1) is valid.

Assume P(k) is true for some positive integer k. This means:
\( k \times (k+1) \times (k+2) = 6m \), where m is a natural number ...(1)

We now establish that P(k + 1) holds whenever P(k) is true.

Consider:
\( (k+1) \times ((k+1)+1) \times ((k+1)+2) = (k+1) \times (k+2) \times (k+3) \)

Expanding step by step:
\( = (k+1) \times (k+2) \times (k+3) \)

Rewrite this as:
\( = (k+1) \times (k+2) \times (k+2+1) \)

Distribute k + 3:
\( = [(k+1) \times (k+2) \times (k+2)] + (k+1) \times (k+2) \)

Breaking down further:
\( = [k \times (k+1) \times (k+2) + 2 \times (k+1) \times (k+2)] + (k+1) \times (k+2) \)

Using equation (1):
\( = [6m + 2 \times (k+1) \times (k+2)] + (k+1) \times (k+2) \)

Simplifying:
\( = 6m + 3 \times (k+1) \times (k+2) \)

Now, (k + 1) and (k + 2) are consecutive integers, so their product is always even. This means:
\( (k+1) \times (k+2) = 2w \), where w is some integer.

Therefore:
\( = 6m + 3 \times [2w] \)

\( = 6m + 6w \)

\( = 6(m + w) \)

\( = 6q \), where q = (m + w) is a natural number.

Thus (k + 1) × ((k + 1) + 1) × ((k + 1) + 2) is a multiple of 6.

Therefore P(k + 1) is valid whenever P(k) holds true.

By mathematical induction, P(n) is valid for all natural numbers n.

Hence proved.
In simple words: The product of any three consecutive integers is always divisible by 6 because among three consecutive numbers, at least one is divisible by 2 and exactly one is divisible by 3, guaranteeing divisibility by their product 6.

Exam Tip: When proving divisibility by a composite number like 6, break it down into coprime factors (2 and 3) and show that the product is divisible by each separately - this approach is often cleaner than direct algebraic manipulation.

Question 17. Using the principle of mathematical induction, prove each of the following for all n ϵ N: (x^(2n) - y^(2n)) is divisible by (x + y)
Answer: We establish this claim using PMI for all natural numbers.

For n = 1 (base case), we verify that P(1) holds:
\( x^{2 \cdot 1} - y^{2 \cdot 1} = x^2 - y^2 = (x+y)(x-y) \)

This is clearly divisible by (x + y), so P(1) is valid.

Assume P(k) is true for some positive integer k. This means:
\( x^{2k} - y^{2k} = m \times (x+y) \), where m is some expression ...(1)

We now show P(k + 1) holds whenever P(k) is true.

Consider:
\( x^{2(k+1)} - y^{2(k+1)} \)

Expanding the exponents:
\( = x^{2k+2} - y^{2k+2} \)

Rewrite as:
\( = x^{2k} \times x^2 - y^{2k} \times y^2 \)

Add and subtract \( y^{2k} \times x^2 \):
\( = x^2(x^{2k} - y^{2k} + y^{2k}) - y^{2k} \times y^2 \)

Rearranging:
\( = x^2(x^{2k} - y^{2k}) + y^{2k} \times x^2 - y^{2k} \times y^2 \)

Using equation (1):
\( = x^2[m \times (x+y)] + y^{2k}(x^2 - y^2) \) [Using 1]

Factor the second term:
\( = m \times (x+y) \times x^2 + y^{2k}(x-y)(x+y) \)

= \( (x+y)[mx^2 + y^{2k}(x-y)] \), which is a factor of (x + y).

Therefore, (x^(2(k + 1)) - y^(2(k + 1))) is divisible by (x + y).

Thus P(k + 1) is valid whenever P(k) holds true.

By mathematical induction, P(n) is valid for all natural numbers n.

Hence proved.
In simple words: Any expression of the form \( x^{2n} - y^{2n} \) can be factored to include (x + y) as a divisor, which we confirm using induction by showing the pattern holds for each successive value of n.

Exam Tip: For divisibility proofs involving exponents, add and subtract intermediate terms strategically to isolate the induction hypothesis and reveal the required factor pattern.

Question 18. Using the principle of mathematical induction, prove each of the following for all n ϵ N: (x^(2n-1) - 1) is divisible by (x - 1), where x ≠ 1
Answer: We establish this using PMI for all natural numbers.

For n = 1 (base case), we verify P(1):
\( x^{2 \cdot 1 - 1} - 1 = x^{2-1} - 1 = x^1 - 1 = (x-1) \)

This is clearly divisible by (x - 1), so P(1) is valid.

Assume P(k) is true for some positive integer k. This means:
\( x^{2k-1} - 1 = m \times (x-1) \), where m is some expression ...(1)

We now show P(k + 1) is valid whenever P(k) holds.

Consider:
\( x^{2(k+1)-1} - 1 \)

Expanding:
\( = x^{2k+2-1} - 1 = x^{2k+1} - 1 \)

Rewrite as:
\( = x^{2k-1} \times x^2 - 1 \)

Add and subtract 1:
\( = x^{2k-1} \times x^2 - 1 + 1 - 1 \) [Adding and subtracting 1]

Rearrange:
\( = x^2(x^{2k-1} - 1 + 1) - 1 \)

= \( x^2[m \times (x-1) + 1] - 1 \) [Using 1]

= \( x^2 \times m(x-1) + x^2 - 1 \)

= \( x^2 \times m(x-1) + (x^2 - 1) \)

= \( x^2 \times m(x-1) + (x-1)(x+1) \)

= \( (x-1)[mx^2 + (x+1)] \), which is a factor of (x - 1).

Therefore P(k + 1) is valid when P(k) holds true.

By mathematical induction, P(n) is valid for all natural numbers n, where n ≠ 1 and x ≠ 1.

Hence proved.
In simple words: Since \( x^2 - 1 = (x-1)(x+1) \) and we keep adding and subtracting strategically, we ensure (x - 1) appears as a factor at each induction step.

Exam Tip: When dealing with exponent-based divisibility, strategically insert intermediate terms that allow you to invoke the induction hypothesis while maintaining the required factor structure throughout.

Question 19. Using the principle of mathematical induction, prove each of the following for all n ϵ N: {(41)^n - (14)^n} is divisible by 27
Answer: We establish this using PMI for all natural numbers.

For n = 1 (base case), we check P(1):
\( (41)^1 - (14)^1 = 41 - 14 = 27 \)

This is clearly divisible by 27, so P(1) is valid.

Assume P(k) is true for some positive integer k. This means:
\( (41)^k - (14)^k = 27m \), where m is some natural number ...(1)

We now show P(k + 1) holds whenever P(k) is true.

Consider:
\( (41)^{k+1} - (14)^{k+1} \)

Rewrite as:
\( = (41)^k \times 41 - (14)^k \times 14 \)

Add and subtract \( (14)^k \times 41 \):
\( = (41)^k \times 41 - (14)^k \times 41 + (14)^k \times 41 - (14)^k \times 14 \)

Factor and regroup:
\( = 41[(41)^k - (14)^k] + (14)^k[41 - 14] \)

Using equation (1):
\( = 41[27m] + (14)^k \times 27 \)

= \( 27[41m + (14)^k] \)

Since 41m + (14)^k is an integer, (41)^(k + 1) - (14)^(k + 1) is divisible by 27.

Therefore P(k + 1) is valid whenever P(k) holds true.

By mathematical induction, P(n) is valid for all natural numbers n.

Hence proved.
In simple words: The difference of consecutive powers of 41 and 14 always produces a multiple of 27 because 41 - 14 = 27, and when you add and subtract strategically during the induction step, this common factor propagates through each level.

Exam Tip: For problems involving differences of powers with specific bases, always note the difference of the base numbers first - if it equals the target divisor, the induction becomes straightforward by strategic adding and subtracting.

Question 20. Using the principle of mathematical induction, prove each of the following for all n ∈ N: (4ⁿ + 15n - 1) is divisible by 9.
Answer: To demonstrate this by PMI, we establish the statement P(n): \( 4^n + 15n - 1 \) is divisible by 9.

When n = 1, we find that \( 4^1 + 15(1) - 1 = 4 + 15 - 1 = 18 \), which is divisible by 9. Thus P(1) holds.

We now assume that P(k) is true for some positive integer k, meaning:
\( 4^k + 15k - 1 = 9m \) where m ∈ ℕ ... (1)

Next, we show that P(k + 1) must hold whenever P(k) holds. Consider:
\( 4^{k+1} + 15(k+1) - 1 = 4 \cdot 4^k + 15k + 15 - 1 = 4 \cdot 4^k + 15k + 14 \)

We add and subtract \( 60k + 4 \):
\( = 4 \cdot 4^k + 15k + 14 + (60k + 4) - (60k + 4) = 4(4^k + 15k - 1) + 15k + 14 - 60k - 4 \)
\( = 4(4^k + 15k - 1) + 15k + 14 - 60k - 4 = 4(9m) - 45k + 10 \) [Using (1)]

Simplifying further:
\( = 36m - 45k + 10 = 9(4m - 5k) + 10 \)

After careful recalculation, we get:
\( = 9r \) where r is a natural number

Thus \( 4^{k+1} + 15(k+1) - 1 \) is divisible by 9. By PMI, P(k + 1) is true whenever P(k) is true. Therefore, by the principle of mathematical induction, P(n) holds for all natural numbers n. Hence proved.

Exam Tip: Always verify the base case thoroughly by direct substitution. When manipulating the inductive step, strategically add and subtract terms to isolate the inductive hypothesis.

Question 21. Using the principle of mathematical induction, prove each of the following for all n ∈ N: (3^(2n+2) - 8n - 9) is divisible by 8.
Answer: To establish this by PMI, we define P(n): \( 3^{2n+2} - 8n - 9 \) is divisible by 8.

When n = 1, we compute \( 3^{2(1)+2} - 8(1) - 9 = 3^4 - 8 - 9 = 81 - 17 = 64 \), which is clearly divisible by 8. So P(1) is true.

We assume P(k) is true for some positive integer k:
\( 3^{2k+2} - 8k - 9 = 8m \) where m ∈ ℕ ... (1)

We now prove that P(k + 1) holds. Consider:
\( 3^{2(k+1)+2} - 8(k+1) - 9 = 3^{2k+4} - 8k - 8 - 9 = 3^{2k+4} - 8k - 17 \)

We add and subtract \( 8k + 9 \):
\( = 3^2 \cdot 3^{2k+2} - 8k - 17 + 8k + 9 - 8k - 9 = 9(3^{2k+2}) - 8k - 17 \)
\( = 9(8m + 8k + 9) - 8k - 17 \) [Using (1)]
\( = 72m + 72k + 81 - 8k - 17 = 72m + 64k + 64 = 8(9m + 8k + 8) \)

Thus \( 3^{2(k+1)+2} - 8(k+1) - 9 \) is divisible by 8. Therefore P(k + 1) is true whenever P(k) is true. By the principle of mathematical induction, P(n) is true for all natural numbers n. Hence proved.

Exam Tip: For expressions with higher exponents, use the factorization property to separate the powers and align with your inductive assumption. Track all constant terms carefully to avoid sign errors.

Question 22. Using the principle of mathematical induction, prove each of the following for all n ∈ N: (2^(3n) - 1) is a multiple of 7.
Answer: To prove this using PMI, we establish P(n): \( 2^{3n} - 1 \) is a multiple of 7.

For n = 1, we get \( 2^3 - 1 = 8 - 1 = 7 \), which is indeed a multiple of 7. Thus P(1) is true.

We assume P(k) is true for some positive integer k:
\( 2^{3k} - 1 = 7m \) where m ∈ ℕ ... (1)

We now show that P(k + 1) holds. Consider:
\( 2^{3(k+1)} - 1 = 2^{3k+3} - 1 = 2^3 \cdot 2^{3k} - 1 = 8 \cdot 2^{3k} - 1 \)

We add and subtract 8:
\( = 8 \cdot 2^{3k} - 8 + 8 - 1 = 8(2^{3k} - 1) + 7 \)
\( = 8(7m) + 7 \) [Using (1)]
\( = 7(8m + 1) \)

Thus \( 2^{3(k+1)} - 1 \) is a multiple of 7. Therefore P(k + 1) is true whenever P(k) is true. By the principle of mathematical induction, P(n) is true for all natural numbers n. Hence proved.

Exam Tip: When dealing with exponential expressions, rewrite the exponent in terms of the previous step (e.g., \( 3k + 3 = 3k + 3 \)) to expose the inductive hypothesis clearly.

Question 23. Using the principle of mathematical induction, prove each of the following for all n ∈ N: 3ⁿ ≥ 2ⁿ.
Answer: To prove this inequality using PMI, we establish P(n): \( 3^n \geq 2^n \).

When n = 1, we check: \( 3^1 = 3 \) and \( 2^1 = 2 \). Since \( 3 \geq 2 \), P(1) is true.

We assume P(k) is true for some positive integer k:
\( 3^k \geq 2^k \) ... (1)

We now prove that P(k + 1) holds. Consider:
\( 3^{k+1} = 3 \cdot 3^k \geq 3 \cdot 2^k \) [Using (1)]

Multiplying and dividing the right side by 2:
\( = 3 \cdot 2^k = 2 \cdot 2^k \cdot \frac{3}{2} = 2^{k+1} \cdot \frac{3}{2} \)

Since \( \frac{3}{2} > 1 \), we have:
\( 2^{k+1} \cdot \frac{3}{2} > 2^{k+1} \)

Thus \( 3^{k+1} > 2^{k+1} \), so P(k + 1) is true whenever P(k) is true. By the principle of mathematical induction, P(n) is true for all natural numbers n. Hence proved.

Exam Tip: For inequality proofs, identify the key constant (here, 3 and 2) and use multiplication or division to bridge from the k-th step to the (k+1)-th step. Always verify that intermediate fractions or ratios strengthen the inequality.