Question

In the product of expansion of (1 + 2x)⁶ (1 – x)⁷, find the coefficient of x⁵.

Answer 1

Given expression

= (1 + 2x)⁶ (1 – x)⁷

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

= 1 + 6 × 2x + 15 × 4x² + 20 × 8x³
+ 15 × 16x⁴+ 6 × 32 x 5+ 1 × 64x⁶

or (1 + 2x)⁶ = 1 + 12x + 60x² + 160x³
+ 240x⁴+ 192x⁵ + 64x⁶ … (i)

[∵ ⁶C₀ = 1, ⁶C₁ = 6, ⁶C₂ = 15, ⁶C₃ = 20, ⁶C₄ = 15, ⁶C₅ = 6]

and (1 – x)⁷ = ⁷C₀ – ⁷C₁x + ⁷C₂x²– ⁷C₃x³ + ⁷C₄x⁴ – ⁷C₅x⁶ +,..

= 1 – 7x + 21x² – 35x³ + 35x⁴ - 21x⁵ +….. (ii)

[⁷C₀ = 1, ⁷C₁= 7, ⁷C₂ = 21, ⁷C₃ = 35, ⁷C₄= 35, ⁷C₅ = 21]
Now (1 + 2x)⁶ (1 – x)⁷

= [1 + 12x + 60x² + 160x³+ 240x⁴ + 192x⁴+ …]

= [1 – 7x + 21x² – 35x³ + 35x⁴ – 21x⁵ + …]

Coefficient of x⁵ in the above expansion

[- 21 + 12 x 35 – 60 × 35 + 160 × 21+ 240 (-7) + 192 × 1]

= [- 21 + 420 – 2100 + 3360 – 1680 + 192]

= [-3801 + 3972] = 171.