1
$\begingroup$

$\sum_{1}^{\infty } \frac{(n^{2} + n)x^{n-1}}{2^{n-1}}$

I have found convergence interval for series: $\left | x \right | < 2$, but I have no idea how to find the sum of the power series.

$\endgroup$
  • 1
    $\begingroup$ Do you know about differentiating power series term-by-term? (or integrating them) $\endgroup$ – ancientmathematician May 19 '17 at 14:57
2
$\begingroup$

$$\sum\limits_{n=1}^\infty{(n^2+n)x^{n-1}\over2^{n-1}} = 4{d^2\over dx^2}\sum\limits_{n=1}^\infty{x^{n+1}\over2^{n+1}} = 4{d^2\over dx^2}{{x^2\over2^2}\over1-{x\over2}} = 2{d^2\over dx^2}{x^2\over2-x} = 2{d^2\over dx^2}{x^2-4+4\over2-x} = 2{d^2\over dx^2}\left(-x-2 + {4\over2-x}\right) = {8\cdot1\cdot2\over(2-x)^3} = {16\over(2-x)^3}$$

$\endgroup$
4
$\begingroup$

Hint: Recall that $$\frac{1}{1-x} = \sum_{n=0}^{\infty}x^n$$ $$\left(\frac{1}{1-x}\right)' = \sum_{n=1}^{\infty}nx^{n-1}$$ $$\left(\frac{1}{1-x}\right)'' = \sum_{n=2}^{\infty}n(n-1)x^{n-2} = \sum_{n=1}^{\infty}n(n+1)x^{n-1} = \sum_{n=1}^{\infty}(n^2+n)x^{n-1}$$ What if you replace $x$ by $\frac{x}{2}$.

$\endgroup$

Your Answer

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