Find the coefficient of x⁵ in (x + 3)⁸

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 x⁵ occurs in the expansion of (x + 3)⁸,we obtain

Tᵣ ₊ ₁ = ⁸Cᵣ(x)^{8 - r} (3)ʳ

Comparing the exponents of x in x⁵ in (Tᵣ ₊ ₁) we obtain

8 - r = 5

r = 3

Thus, the coefficient of x⁵ is ⁸C₃(3)³

⁸C₃(3)³ = [8! / (3! 5!)] x (3)³

= [8 x 7 x 6 x (5!)]/[3 x (2!) x (5!)] x 27

= 1512

NCERT Solutions Class 11 Maths Chapter 8 Exercise 8.2 Question 1

Find the coefficient of x⁵ in (x + 3)⁸

Summary:

Using binomial theorem, coefficient of x⁵ in (x + 3)⁸ is to be evaluated. We have found that it is equal to 1512