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Let k be a fixed positive integer. The $n^{th}$ derivative of $\frac{1}{x^k-1}$ has the form $\frac{p_n(x)}{(x^k-1)^{n+1}}$ where $p_n(x)$ is a polynomial of degree n with $p_0=1$. Then the value of $p_n(1)$ is

(A) $(n-1)!(-k)^n$

(B) $n!(-k)^{n-1}$

(C) $(n-1)!(-k)^{n-1}$

(D) $n!(-k)^{n}$

My approach is as follow Let $t(x)=(x^k-1)^{-1}$

$t'(x)=-(x^k-1)^{-2}.kx^{k-1}$

$t''(x)=2(x^k-1)^{-3}.k^2x^{2(k-1)}-(x^k-1)^{-2}.k(k-1)x^{k-2}$

I am not able to map it properly

Samar Imam Zaidi
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1 Answers1

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I think there's a minor mistake in the question: $p_n(x)$ does not have degree $n$. You can see this from your examples of $t'(x)$ and $t''(x)$. Anyway, here is a possible answer.

$$\frac{p_{n+1}(x)}{(x^k-1)^{n+2}}=\left(\frac{p_n(x)}{(x^k-1)^{n+1}}\right)'=\frac{p_n'(x)(x^k-1)-p_n(x)(n+1)kx}{(x^k-1)^{n+2}}$$

Hence, $p_{n+1}(x)=p_n'(x)(x^k-1)-p_n(x)(n+1)kx$. Define $a_n=p_n(1)$. Then: $$a_{n+1}=p_{n+1}(1)=-k(n+1)a_n \ \ \textrm{and } \ a_0=1$$

We can prove by induction that $a_n=(-k)^nn!$

  • $n=1$: $a_1=-k(0+1)a_0=-k$
  • Assume $a_m=(-k)^mm!$ for $m\geq 1$. Then $a_{m+1}=-k(m+1)(-k)^mm!=(-k)^{m+1}(m+1)!$

In conclusion, the answer is (D).

Jandro
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