What should be the strategy to find
$$\sum_{k=0}^{n}k^2\binom{n}{k}$$
Can this be done by making a series of $x$ and integrating?
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2Your title says $n^n\cdot\binom{n}{n}$, but your question says $n^2\cdot\binom{n}{n}$. Which do you mean? – Ben Sheller Sep 11 '15 at 16:20
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I am pretty sure OP meant $n^2 \binom{n}{n}$ – Trogdor Sep 11 '15 at 16:23
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@Trogdor But it really could be either one, based on the terms that have been written out. That's why I was waiting to edit until OP responded. – Ben Sheller Sep 11 '15 at 16:24
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I edited the title to fix some formatting issues, but feel free to change the $n^2$ to $n^n$ if that's what you intended. – anomaly Sep 11 '15 at 16:25
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I wanted to write n^2. I should have written one more term – Archisman Panigrahi Sep 11 '15 at 16:27
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... and a few other posts. Please search before asking. – Oct 19 '15 at 11:23
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Interesting question..... I know you'll get the traditional replies, which will probably help you solve the problem. However, if you get time, try to think of it this way:
Consider $n$ players. You want to make a team out of these $n$ players. Note that the team can have any number of players! (This would explain the $^nC_0 + ^nC_1 + ...$.)
Now consider that you have two positions up for take: Captain, and Vice-Captain. You need a player to fill in for those positions. Any player can fill for that position. Not only that, a single player can also occupy both positions! (This is for the coefficient, $0^2 , 1^2 , 2^2 , etc.$)
Now... Think of it in the reverse. For the first case, suppose you have to chose a single person to become captain and vice captain: $n$ choices. Now, we have $n-1$ players to distribute.. So: $2^{n-1}$. Total is: $n2^{n-1}$
For the second case, consider you have to chose two different players for captain and vice-captain. So number of choices: $n(n-1)$. Now we have $n-2$ players remaining to distribute: $2^{n-2}$. Total is: $n(n-1)2^{n-2}$
Thus, the total number of ways of doing something like this is: $$n2^{n-1} + n(n-1)2^{n-2}$$
Gummy bears
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@BrianM.Scott I just left it like that because I thought it would be easier to recognize. But yes, of course. – Gummy bears Sep 12 '15 at 04:30
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Hint
Start with $$(1+x)^n = \sum_{k=0}^n \binom{n}{k}x^k$$ Differentiate, multiply by $x$ and differentiate again. What do you get?
Archisman Panigrahi
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N. S.
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2@N.S.: Not in my experience, except perhaps by people for whom English is a second language. The English verb is differentiate. – Brian M. Scott Sep 11 '15 at 17:34
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@N.S. It's fine. Just seemed strange to me. Conveys the message anyhow so... – Gummy bears Sep 12 '15 at 04:31
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@N.S. How to find $$\sum_{k=0}^{n}k^r\binom{n}{k}$$. Should I make this a new question? In these type of problems, how do you find when to differenciate and when to multiply with $x$? Is it a matter of practicing or something? – Archisman Panigrahi Sep 12 '15 at 12:31
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1@ArchismanPanigrahi The point is that if you don't multiply by $x$ you get $\sum_{k=0}^n k(k-1) \binom{n}{k}$. You can fix this either the way I did, or alternately by splitting $k(k-1)\binom{n}{k}=k^2\binom{n}{k}-k\binom{n}{k}$ and observe that you calculated the second before. – N. S. Sep 13 '15 at 15:55
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1@ArchismanPanigrahi For the general question, you can either alternate between differentiation and multiplication by $x$. Or if you just differentiate $r$ times you get $\sum_{k=0}^{n}k(k-1)(k-2)..(k-r+1)\binom{n}{k}$. If you pen the bracket, you get a recurrence of the form: $$ \sum_{k=0}^{n}k^r\binom{n}{k} -A \sum_{k=0}^{n}k^{r-1}\binom{n}{k}+B \sum_{k=0}^{n}k^{r-2}\binom{n}{k}+..=\mbox{something}$$ I find multiplication easier. – N. S. Sep 13 '15 at 15:57
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Let $$\displaystyle S=\sum^{n}_{k=0}k^2 \binom{n}{k} \;,$$ Now using $\displaystyle \bullet \; \binom{n}{k} = \frac{n}{k}\cdot \binom{n-1}{k-1}$
So we get $$\displaystyle S = \sum^{n}_{k=0}k^2 \cdot \frac{n}{k}\binom{n-1}{k-1}\;,$$ Again using $\displaystyle \bullet \; \binom{n-1}{k-1} = \frac{n-1}{k-1}\cdot \binom{n-2}{k-2}$
So $$\displaystyle S = n\sum^{n}_{k=1}k\binom{n-1}{k-1} = n\sum^{n}_{k=0}[(k-1)+1]\binom{n-1}{k-1}$$
So we get $$\displaystyle S=n\sum^{n}_{k=1}(k-1)\binom{n-1}{k-1}+n\sum^{n}_{k=1} \binom{n-1}{k-1}$$
So we get $$\displaystyle S=n\sum^{n}_{k=2}(k-1)\cdot \frac{n-1}{k-1}\cdot \binom{n-2}{k-2}+n\sum^{n}_{k=1}\binom{n-1}{k-1}$$
So we get $$\displaystyle S=n(n-1)\sum^{n}_{k=0}\binom{n-2}{k-2}+n\sum^{n}_{k=0}\binom{n-1}{k-1}$$
Now Using $\displaystyle \bullet\; \sum^{n}_{k=0}\binom{n}{k} = 2^{k}$
So we get $$\displaystyle S = n(n-1)\cdot 2^{n-2}+n\cdot 2^{n-1}$$
juantheron
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I think the limits should be changed in this too. "So we get $$\displaystyle S=n(n-1)\sum^{n}{k=0}\binom{n-2}{k-2}+n\sum^{n}{k=0}\binom{n-1}{k-1}$$" $ k=2$ to $n$ . Also when we change the lower limit, are not we discarding $k= 0$ and $1$ terms? – Archisman Panigrahi Sep 12 '15 at 12:27