2
$\begingroup$

Find the sum of the power series $\sum\limits_{n=1}^\infty \frac{(n+2)!}{(2!)(n!)}x^n$

frist I use $$\sum_{n=1}^\infty x^n = \frac{x}{1-x}$$

and multiple two side by $x^2$

can get $$\sum_{n=1}^\infty x^{n+2} = \frac{x^3}{1-x}$$

then diff each side two times

we can obtain $$\sum_{n=1}^\infty (n+1)(n+2)x^n= \frac{2x^3-6x^2+6x}{(1-x)^3}$$

but not the solution $=\frac{1}{(1-x)^3}$

$\endgroup$
  • $\begingroup$ use Newton's binomial theorem $\endgroup$ – user379195 May 11 '17 at 16:58
  • $\begingroup$ the result should be $$-\frac{x \left(x^2-3 x+3\right)}{(x-1)^3}$$ $\endgroup$ – Dr. Sonnhard Graubner May 11 '17 at 17:03
  • $\begingroup$ You should be using $\sum_{n=0}^{\infty}x^{n+2}=\frac{x^2}{(1-x)}$ then differentiate and which will get you $\frac{2x-x^2}{(1-x)^2}=n+2 \sum_{n=0}^{\infty}x^{n+1}$ then differentiate again and you'll get your final answer by finally diving by 2. $\endgroup$ – Iti Shree May 11 '17 at 17:06
  • $\begingroup$ thanks,I found that this question is wrong, n is start from zero. $\endgroup$ – Andrew Hung May 11 '17 at 17:27
1
$\begingroup$

Starting from zero,

$\begin{array}\\ \sum\limits_{n=0}^\infty \frac{(n+2)!}{(2!)(n!)}x^n &=\sum\limits_{n=0}^\infty \binom{n+2}{2}x^n\\ &=\frac1{x^2}(1-x)^{-3}\\ \end{array} $

by the generalized binomial theorem (since $\binom{-n}{k} =(-1)^n \binom{n+k-1}{k} $).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.