How can I evaluate the sum $\sum_{k=1}^{n} k^2 2^k$?

Question

I am trying to evaluate the sum $$ \sum_{k=1}^{n} k^2 2^k. $$ I was following this post and came up with this: \begin{align*} S(n) &= \sum_{k=1}^{n} k^2 2^k \\ S(n+1) &= S(n) + \sum_{k=2}^{n+1} k^2 2^k \\ &= 2 + 2\sum_{k=1}^{n} k^2 2^k \\ &= 2 + \sum_{k=1}^{n} (k+1)^2 2^{k+1} \\ &= 2 + 2\sum_{k=1}^{n} 2^k(k^2 + 2k + 2) \\ &= 2 + 2 \left( \sum_{k=1}^{n} k^2 2^k + 2\sum_{k=1}^{n}k2^k + 2\sum_{k=1}^{n} 2^k \right) \\ &= 2 + 2 \left( S(n) + 2\sum_{k=1}^{n}k2^k + 2\sum_{k=1}^{n} 2^k \right) \end{align*}

And this is where I'm stuck. I plugged it into WolframAlpha and apparently the answer is $S(n) = 2(2^nn^2 - 2^{n+1}n+3\cdot 2^n-3)$. Any help would be appreciated.

Answer 1

Hint.

As

$$ S_n(x) = \frac{x^{n+1}-1}{x-1} $$

we have

$$ \sum_{k=0}^n k^2x^k = x\frac{d}{dx}\left(x\frac{d}{dx}S(x)\right) $$

Answer 2

$\sum_{k=0}^nk^2 x^k=x\sum_{k=0}^nk^2x^{k-1}$

$=x\frac{d}{dx}\sum_{k=0}^nkx^k=x\frac{d}{dx}x\sum_{k=0}^nkx^{k-1}$

$=x\frac{d}{dx}x\frac{d}{dx}\sum_{k=0}^nx^k$

$=x\frac{d}{dx}x\frac{d}{dx}\frac{x^{n+1}-1}{x-1}$

I'll let you do the calculus and evaluate the expression for $x=2$.

Note that I started the sum at $k=0$, but it doesn't matter since the added term is $0$.

Answer 3

If you insist on using the method you provided in the link, then a corrected derivation would look something like: \begin{align} S(n+1) &= \sum_{k=1}^{n+1} k^22^k \\ &= 2 + \sum_{k=2}^{n+1} k^22^k \\ &= 2 + \sum_{i=1}^n (i+1)^22^{i+1} \\ &= 2 + 2\left(\sum_{i=1}^n i^22^i + 2\sum_{i=1}^ni2^i + \sum_{i=1}^n2^i\right) \\ &= 2 + 2\left(S(n) + 2\sum_{i=1}^ni2^i + \sum_{i=1}^n2^i\right). \end{align} Then by using the fact that $S(n+1) = S(n) + (n+1)^22^{n+1}$ we can solve for $S(n)$. You may recognise $\sum_{i=1}^n2^i$ as the finite sum of a geometric series (this has a standard formula). The sum $\sum_{i=1}^ni2^i$ is similar in form to our original problem and thus a similar argument can be used (i.e. we take $A(n) = \sum_{i=1}^ni2^i$ and proceed similarly).

I would not use this method myself and instead use the method demonstrated in the other answers.

Answer 4

An easy trick.

Consider $$S_n=\sum_{k=1}^n k^2 x^k=\sum_{k=1}^n (k(k-1)+k) x^k=x^2\sum_{k=1}^n k(k-1) x^{k-2}+x\sum_{k=1}^n k x^{k-1}$$ that is to say $$S_n=x^2\left(\sum_{k=1}^n x^{k} \right)''+x\left(\sum_{k=1}^n x^{k} \right)'$$