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Show that ... $$\sum_{n=1}^\infty \frac{n^2} {2^n} = 6$$

How would one go about in showing that the above power series equals to 6? I would assume that it has something to do with the following properties such as ...

f(x) = $\sum {Cn}·{X^n}$, g(x) = $\sum {Dn}·{X^n}$

  1. f(x) ± g(x) = $\sum_{n=b}^\infty {Cn}·{X^n}$ ± $\sum_{n=b}^\infty {Dn}·{X^n}$ = $\sum_{n=b}^\infty ({Cn ± Dn})·X^n$

  2. f $(K · {X^m})$ = $\sum_{n=b}^\infty {Cn}({K·{X^m}})^n$

  3. $\sum_{n=b}^\infty {Cn} {X^{n+m}} $ = ${X^m} \sum_{n=b}^\infty {Cn}·{X^n}$

  4. if f(x) = $\sum_{n=0}^\infty {Cn}·{X^n}$ then f'(x) = $\sum_{n=1}^\infty {Cn}·{nX^{n-1}}$

  5. if f(x) = $\sum_{n=0}^\infty {Cn}·{X^n}$ then $\int f(x)$ = $\sum_{n=0}^\infty$ $\frac {{Cn}·X^{n+1}} {n+1}$

  6. $\sum_{n=b}^\infty {Cn}·{X^n}$ is continuous on Interval of Convergence.

This is the completed work so far ...

$$\sum_{n=1}^\infty \frac{n^2} {2^n} = \sum_{n=1}^\infty {(1/2)^n} ·{n^2}$$

let f(x) = $$ \sum_{n=1}^\infty {X^n}{n^2}$$

then $$ \int f(x) = \sum_{n=1}^\infty \frac {X^{n+1}} {n+1} · {n^2}$$

I assume the next step should be to cancel ${n^2}$ and n+1, but I'm not quite getting the connection. How would you proceed with this?

LeviRivai
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2 Answers2

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Hinte: Try starting with $\sum_{n=0}^\infty x^n = {1 \over 1-x}$ for $|x|<1$ and differentiate both sides. Multiply the result by $x$ and repeat. Adjust as necessary. Rinse & dry.

copper.hat
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Consider$$S_k=\sum_{n=1}^k {n^2} x^n=\sum_{n=1}^k [n(n-1)+n] x^n=\sum_{n=1}^k n(n-1) x^n+\sum_{n=1}^k n x^n$$ that is to say $$S_k=x^2\sum_{n=1}^k n(n-1) x^{n-2}+x\sum_{n=1}^k n x^{n-1}=x^2\left(\sum_{n=1}^k x^{n} \right)''+x\left(\sum_{n=1}^k x^{n} \right)'$$

Claude Leibovici
  • 289,558