If n is an odd positive integer and (1 + x + x^2 + x^3)^n = ∑ (r =0 to 3n) ar x^r , then a0 – a1 + a2 – a3 + ..... – a3n equals

Question

If n is an odd positive integer and (1 + x + x² + x³)ⁿ = \(\sum_{r=0}^{3n} a_r \,x^r\), then a₀ – a₁ + a₂ – a₃ + ..... – a_3n equals

(a) –1 

(b) 1 

(c) 4ⁿ 

(d) 0

Answer 1

Answer : (d) 0

Given, (1 + x + x² + x³)ⁿ = \(\sum_{r=0}^{3n} a_r\, x^r\)

⇒ (1 + x + x² + x³)ⁿ = a₀ + a₁x + a₂x² + ..... + a_3n x³ⁿ

Putting x = –1 in the above-given equation, we have

a₀ – a₁ + a₂ – a₃ + ..... – a_3n = 0