0

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.

Lukas Zamora
  • 23
  • 3
    $S(n+1)\ne S(n)+\sum_{k=2}^{n+1}k^22^k$, so... – Simply Beautiful Art Sep 03 '19 at 00:01
  • 2
    The state-of-the-art trick is by differentiating the polynomial $f(x) = \sum_{k = 0}^n x^k = \frac{1 - x^{n + 1}}{1 - x}$ for several times. – Zhanxiong Sep 03 '19 at 00:08
  • Although there are some mistakes on the first lines, your fourth line mysteriously is correct. My hint for you is: you probably know how to compute $\sum_{k=1}^n 2^k$. A next step would be to use the strategy above to find an expression for $\sum_{k=1}^n k \cdot 2^k$. From there, you can use your final expression. – D. Ungaretti Sep 03 '19 at 00:08
  • See https://math.stackexchange.com/questions/1688659/induction-proof-k2-times-2k – Gerry Myerson Sep 03 '19 at 00:09
  • @GerryMyerson Nicely spotted, but this is not the same. No induction here. – amsmath Sep 03 '19 at 00:11
  • @ams, the title of the question here is, "How can I evaluate the sum $\sum_{k=1}^nk^22^k$?" The link answers the question in the title. – Gerry Myerson Sep 03 '19 at 00:18
  • @GerryMyerson I'm assuming OP would like to know how to get there without knowing the result beforehand. – amsmath Sep 03 '19 at 00:21
  • @ams, you may be right. OP should edit the title in such a way as to reflect what he/she really wants to know. – Gerry Myerson Sep 03 '19 at 00:23

4 Answers4

0

Hint.

As

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

we have

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

Cesareo
  • 33,252
0

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

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

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

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

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 $0$.

herb steinberg
  • 12,765
0

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.

Timothy Hedgeworth
  • 290
0

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 $$S_n=x^2\left(\sum_{k=1}^n x^{k} \right)''+x\left(\sum_{k=1}^n x^{k} \right)'$$

Claude Leibovici
  • 260,315