NCERT Class 11 Maths Binomial Theorem

8.1 Introduction
In earlier classes, we have learnt how to find the squares and cubes of binomials like a + b and a – b. Using them, we could evaluate the numerical values of numbers like (98)2 = (100 – 2)2, (999)3 = (1000 – 1)3, etc. However, for higher powers like (98)5, (101)6, etc., the calculations become difficult by using repeated multiplication. This difficulty was overcome by a theorem known as binomial theorem. It gives an easier way to expand (a + b)n, where n is an integer or a rational number. In this Chapter, we study binomial theorem for positive integral indices only.

8.2 Binomial Theorem for Positive Integral Indices

Let us have a look at the following identities done earlier:

(a+ b)⁰ = 1       a + b ≠ 0

(a+ b)¹ = a + b

(a+ b)² = a² + 2ab + b²

(a+ b)³ = a³ + 3a2b + 3ab2 + b³

(a+ b)⁴ = (a + b)³ (a + b) = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴

In these expansions, we observe that

(i) The total number of terms in the expansion is one more than the index. For example, in the expansion of (a + b)2 , number of terms is 3 whereas the index of (a + b)² is 2.

(ii) Powers of the first quantity ‘a’ go on decreasing by 1 whereas the powers of the second quantity ‘b’ increase by 1, in the successive terms.

(iii) In each term of the expansion, the sum of the indices of a and b is the same and is equal to the index of a + b

Please refer to attached file for NCERT Class 11 Maths Binomial Theorem