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This question was asked in my analysis quiz and I had no clue in the exam how it could be solved.So, I am asking here .

Find the sum of series $\sum_{n=0}^{\infty} \frac{n^2} {2^{n} } $ .

Unfortunately, I have no idea on how to approach this problem.

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    Start with the geometric series $\sum_{n=0}^{\infty} r^n=(1-r)^{-1}$ – Integrand Jul 01 '20 at 17:59
  • @Integrand, do you mean, using r=1/2? – Ashley Jul 01 '20 at 18:02
  • Answered here: https://math.stackexchange.com/q/955758/42969, and here: https://math.stackexchange.com/q/593996/42969, and here: https://math.stackexchange.com/q/338852/42969 – all found with Approach0 – Martin R Jul 01 '20 at 18:12
  • Too many questions begin or end with "I don't even know how to begin with this problem". While this may be true [...], it is still not a valid reason to limit your post to the statement of the problem without any mention of your own thoughts. – From Avoid "no clue" questions. – Martin R Jul 01 '20 at 18:16

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We start with $$\sum_{n=0}^\infty \frac{x^n}{2^n} = \frac{2}{2-x}.$$ Taking the derivative, we see that $$\sum_{n=1}^\infty \frac{nx^{n-1}}{2^n} = \frac{2}{(2-x)^2}.$$ Multiplying by $x$ and taking the derivative once more, we see that $$\sum_{n=1}^\infty \frac{n^2 x^{n-1}}{2^n} = \frac{4+2x}{(2-x)^3}.$$ Lastly, plugging in $x=1$ gives that $$\sum_{n=1}^\infty \frac{n^2}{2^n} = 6.$$

Red Sleuth
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