Find the coefficient of x⁵ in the product (1 + 2x)⁶ (1 – x)⁷ using binomial theorem

Solution:

We will find the expansions of (1 + 2x)⁶ and (1 – x)⁷ using the binomial theorem.

(1 + 2x)⁶ = ⁶C₀ (1)⁶ + ⁶C₁ (1)⁵ (2x) + ⁶C₂ (1)⁴ (2x)² + ⁶C₃ (1)³ (2x)³ + ⁶C₄ (1)² (2x)⁴ + ⁶C₅ (1)(2x)⁵ + ⁶C₆(2x)⁶

= 1 (1) + 6 (2x) + 15 (2x)² + 20 (2x)³ + 15 (2x)⁴ + 6 (2x)⁵ + 1 (2x)⁶

= 1 +12 x + 60x² + 160x³ + 240x⁴ + 192x⁵ + 64x⁶

(1 – x)⁷ = ⁷C₀ (1)⁷ - ⁷C₁ (1)⁶ (x) + ⁷C₂ (1)⁵ (x)² - ⁷C₃ (1)⁴ (x)³ + ⁷C₄ (1)³ (x)⁴ - ⁷C₅ (1)² (x)⁵ + ⁷C₆ (1) (x)⁶ + ⁷C₇ x⁷

= 1 - 7x + 21x² - 35x³ + 35x⁴ - 21x⁵ + 7x⁶ - x⁷

Now we will find the given product.

(1 + 2x)⁶ (1 – x)⁷ = (1 +12 x + 60x² + 160x³ + 240x⁴ + 192x⁵ + 64x⁶) (1 - 7x + 21x² - 35x³ + 35x⁴ - 21x⁵ + 7x⁶ - x⁷)

We have to find the coefficient of x⁵ in this product.

Since it is difficult to find the entire product, we will find only coefficients of x⁵ and add them

Coefficient of x⁵ = 1(-21) + 12 (35) + 60 (-35) + 160 (21) + 240 (-7) + 192 (1) = 171

NCERT Solutions Class 11 Maths Chapter 8 Exercise ME Question 3

Find the coefficient of x⁵ in the product (1 + 2x)⁶ (1 – x)⁷ using binomial theorem

Summary:

The coefficient of x⁵ in the product (1 + 2x)⁶ (1 – x)⁷ using binomial theorem is 171