Question

Show that the coefficient of x⁵ in the expansion of product (1 + 2x)⁶ (1 – x)⁷ is 171.

Answer 1

We have (1 + 2x) ⁶ (1 – ) ⁷ 

= [1 + ⁶C₁ (2x) + ⁶C₂ (2x)² + ⁶C₃ (2x)³ + ⁶C₄ (2x)⁴ + ⁶C₅ (2x) 5 + ⁶C₆ (2x) 6 ] × [1 – ⁷C₁ x + ⁷C₂ x 2 – ⁷C₃x³ + ⁷C₄ x ⁴ – ⁷C₅ x⁵ + ⁷C₆ x⁶ –⁷C₇ x⁷ ] ÷ (1 + 12x + 60x² + 160x³ + 240x ⁴ + 192x⁵ + 64x⁶ ) × (1 – 7 x + 21x² – 35x³ + 35x⁴ – 21x5 +x⁶ –x⁷ ) 

∴ Coefficients of 5 in the product

 = 1 × (– 21) + 12 × 35 + 60 × (– 35) + 160 × 21 + 240 × (– 7) + 192 × 1 

= – 21 + 420 – 2100 + 3360 – 1680 + 192 

= 171