If n is an odd positive integer and (1 + x + x2 + x3)n = \(\sum_{r=0}^

1.	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 + ..... – a3n equals(a) –1 (b) 1 (c) 4n (d) 0
Answer» 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

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 + ..... – a3n equals(a) –1 (b) 1 (c) 4n (d) 0

Answer»

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