Question

If n ÎN and (1 + x + x² + …. + x^p)ⁿ = $\sum_{r = 0}^{np}{a_{r}x^{r}}$, if 0 £ r £np and r is not a multiple of p + 1 then,

ⁿC₀a_r – ⁿC₁a_r–1 + ⁿC₂a_r–2 – ... + (–1)^r a₀ⁿC_r is –

• A. 3ⁿ
• B. 2ⁿ
• C. 0
• D. None of these

Answer 1

Answer: (C)

0

Answer 2

We have,

(1 – x)ⁿ = ⁿC₀ – ⁿC₁x + ⁿC₂ x² – ⁿC₃ x³ + … + (–1)^{n n}C_nxⁿ and,

(1 + x + x² + ….. + x^p)ⁿ = (a₀ + a₁x + a₂x² + ….. + a_npx^np)

\ (1 – x)ⁿ (1 + x + x² + ….. + x^p)ⁿ

= (ⁿC₀ – ⁿC₁x + ⁿC₂ x² – ⁿC₃ x³ + …. + (–1)^{n n}C_n xⁿ)

× (a₀ + a₁x + a₂x² + ….. + a_np x^np) …(1)

Now, Coefficient of x^r on R.H.S. of (1)

= a_r ⁿC₀ – a_r–1 ⁿC₁ + a_r–2 ⁿC₂ –…. + a₀(–1)^{r n}C_r

LHS of (1) = (1 – x)ⁿ (1 + x + x² + ….+ x^p)

= (1 – x)ⁿ$\left\{ \frac{1 - x^{p + 1}}{1 - x} \right\}^{n}$

= (1 – x^p+1)ⁿ

Since r is not a multiple of (p + 1). Therefore, the expansion of (1 – x^p+1)ⁿ does not contain x^r in any term.

\ Coefficient of x^r on LHS of (1) = 0

Hence,

ⁿC₀ a_r – ⁿC₁a_r–1 + ⁿC₂ a_r–2 –…..+(–1)^{r n}C_r a₀ = 0.

If r is not a multiple of (p + 1)

Hence (3) is correct answer.