## Putnam, 1962: Summing Binomials

### Putnam, 1962: Summing Binomials

Given, $n \in \mathbb{N}$, show that,

\[\displaystyle\sum^{n}_{r=1}r^{2}\binom{n}{r}=n(n+1)2^{n-2}\]

\[\displaystyle\sum^{n}_{r=1}r^{2}\binom{n}{r}=n(n+1)2^{n-2}\]

Last edited by sowmitra on Tue Sep 04, 2012 9:37 pm, edited 1 time in total.

### Re: Putnam, 1962: Summing Binomials

$\displaystyle\sum_{r=1}^{n}r^{2}\binom{n}{r}=\displaystyle\sum_{r=1}^{n}nr\binom{n-1}{r-1}$ [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.

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.

Last edited by SANZEED on Fri Sep 07, 2012 11:15 pm, edited 1 time in total.

$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$

### Re: Putnam, 1962: Summing Binomials

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

### Re: Putnam, 1962: Summing Binomials

Sorry, I have edited the typo.

$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$

### Re: Putnam, 1962: Summing Binomials

using second derivative can give a nice proof.

*হার জিত চিরদিন থাকবেই*

তবুও এগিয়ে যেতে হবে.........

বাধা-বিঘ্ন না পেরিয়ে

বড় হয়েছে কে কবে.........তবুও এগিয়ে যেতে হবে.........

বাধা-বিঘ্ন না পেরিয়ে

বড় হয়েছে কে কবে.........

### Re: Putnam, 1962: Summing Binomials

And here is the proof:sm.joty wrote:using second derivative can give a nice 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.

$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$

- Tahmid Hasan
**Posts:**665**Joined:**Thu Dec 09, 2010 5:34 pm**Location:**Khulna,Bangladesh.

### Re: Putnam, 1962: Summing Binomials

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 $2^{n-2}(2n+n^2-n)=n(n+1)2^{n-2}$.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.

বড় ভালবাসি তোমায়,মা

### Re: Putnam, 1962: Summing Binomials

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

One one thing is neutral in the universe, that is $0$.

### Re: Putnam, 1962: Summing Binomials

Now, find the sum \[\sum_{i=0}^n\binom nii^3\]

The same trick should work.

The same trick should work.

One one thing is neutral in the universe, that is $0$.