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Given, $n \in \mathbb{N}$, show that,
\[\displaystyle\sum^{n}_{r=1}r^{2}\binom{n}{r}=n(n+1)2^{n-2}\]

Lemma:

$\displaystyle\sum_{r=1}^{n}r^{2}\binom{n}{r}=\displaystyle\sum_{r=1}^{n}nr\binom{n-1}{r-1}$ [Since $\binom{n}{r}=\frac{n}{r}\binom{n-1}{r-1}$]$=n\displaystyle\sum_{r=1}^{n}r\binom{n-1}{r-1}=n\displaystyle\sum_{r=1}^{n}(r-1)\binom{n-1}{r-1}+n\displaystyle\sum_{r=1}^{n}\binom{n-1}{r-1}$
Which is equal to $n\displaystyle\sum_{i=0}^{n-1}i\binom{n}{i}+\displaystyle\sum_{i=0}^{n-1}\binom{n-1}{i}=n\times (n-1)\times 2^{n-2}+2^{n-1}$ and the desired result follows.

SANZEED wrote:$\displaystyle\sum_{r=1}^{n}r^{2}\binom{n}{r}=\displaystyle\sum_{r=1}^{n}nr\binom{n-1}{r-1}$ [Since $\frac{n}{r}\binom{n}{r}=\binom{n-1}{r-1}$]

I think, it is, $\displaystyle\binom{n}{r}=\frac{n}{r}\binom{n-1}{r-1}$

Sorry, I have edited the typo.

using second derivative can give a nice proof.

sm.joty wrote:using second derivative can give a nice proof.

And here is the proof:
From the binomial theorem,we have,$(x+y)^{n}=\displaystyle\sum_{r=0}^{n}\binom{n}{r}x^{n-r}y^{r}$. Substitute $x=1$, then $(1+y)^{n}=\displaystyle\sum_{r=0}^{n}\binom{n}{r}y^{r}$. Now taking the second derivative on each side, we have that $n(n-1)(1+y)^{n-2}=\displaystyle\sum_{r=0}^{n}r(r-1)\binom{n}{r}y^{r-2}\Rightarrow n(n-1)(1+y)^{n-2}=\displaystyle\sum_{r=0}^{n}r^{2}\binom{n}{r}y^{r-2}-\displaystyle\sum_{r=0}^{n}r\binom{n}{r}y^{r-2}$. Substitute $y=1$ and use the lemma $\displaystyle\sum_{r=0}^{n}r\binom{n}{r}=n\cdot 2^{n-1}$ (Which can also be proved taking first derivative instead of second). The result then follows.

Masum wrote: The Instructor and the captain same- we have $2^{n-2}*n^2$ (why?). And if not, $2^{n-2}n$. And the total number is just their sum.

I beg to differ. When the instructor and captain is the same there are $n.2^{n-1}$ choices and when they are different there are $\binom{n}{2}.2!.2^{n-2}$ choices. Since the cases are disjoint, the total number of choices are $2^{n-2}(2n+n^2-n)=n(n+1)2^{n-2}$.

Yeah, you are right. Something was wrong, when I was writing the post.

Now, find the sum \[\sum_{i=0}^n\binom nii^3\]
The same trick should work.