Find the coefficient of a⁵b⁷ in (a - 2b)¹²

Solution:

It is known that (r + 1)^th term, (Tᵣ ₊ ₁) in the binomial expansion of (a + b)ⁿ is given by Tᵣ ₊ ₁ = ⁿCᵣ aⁿ ⁻ ʳ bʳ.

Assuming a⁵b⁷ occurs in the expansion of (a - 2b)¹², we obtain

Tᵣ ₊ ₁ = ¹²Cᵣ(a)¹² ⁻ ʳ (- 2b)ʳ

= ¹²Cᵣ(- 2)ʳ (a)¹² ⁻ ʳ (b)ʳ

Comparing the exponents of x in a⁵b⁷ in (Tᵣ ₊ ₁) we obtain

12 - r = 5 (or) r = 7

From each of these, we get r = 7

Thus, the coefficient of a⁵b⁷ is ¹²C₇(- 2)⁷.

¹²C₇(- 2)⁷ = [12!/(7! 5!)] x (- 2)⁷

= [12 x 11 x 10 x 9 x 8 x (7!)]/[(7!) (5!)] x (- 128)

= - (792) x (128)

= - 101376

NCERT Solutions Class 11 Maths Chapter 8 Exercise 8.2 Question 2

Find the coefficient of a⁵b⁷ in (a - 2b)¹²

Summary:

Using the binomial theorem, the coefficient of a⁵b⁷ in (a - 2b)¹² is to be evaluated. We have found that it is equal to - 101376