calculate-n-0-n-2-3-n-

Question Number 147862 by mathmax by abdo last updated on 24/Jul/21

calculate \sum_{n = 0}^{\infty} \frac{n^{2}}{3^{n}}

Answered by qaz last updated on 24/Jul/21

\underset{n = 0}{\sum^{\infty}} \frac{n^{2}}{3^{n}} = {(xD)}^{2} ∣_{x = \frac{1}{3}} \underset{n = 0}{\sum^{\infty}} x^{n} = (x^{2} D^{2} + xD) ∣_{x = \frac{1}{3}} \frac{1}{1 - x}

= {[\frac{{2 x}^{2}}{{(1 - x)}^{3}} + \frac{x}{{(1 - x)}^{2}}]}_{x = \frac{1}{3}} = \frac{x^{2} + x}{{(1 - x)}^{3}} ∣_{x = \frac{1}{3}} = \frac{3}{2}

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