## Putnam, 1962: Summing Binomials

For discussing Olympiad Level Combinatorics problems
sowmitra
Posts: 155
Joined: Tue Mar 20, 2012 12:55 am

### Putnam, 1962: Summing Binomials

Given, $n \in \mathbb{N}$, show that,
$\displaystyle\sum^{n}_{r=1}r^{2}\binom{n}{r}=n(n+1)2^{n-2}$
Last edited by sowmitra on Tue Sep 04, 2012 9:37 pm, edited 1 time in total.
"Rhythm is mathematics of the sub-conscious."
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sowmitra
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Joined: Tue Mar 20, 2012 12:55 am

### Re: Putnam, 1962: Summing Binomials

Lemma:
"Rhythm is mathematics of the sub-conscious."
Some-Angle Related Problems;

SANZEED
Posts: 550
Joined: Wed Dec 28, 2011 6:45 pm

### 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.
Last edited by SANZEED on Fri Sep 07, 2012 11:15 pm, edited 1 time in total.
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sowmitra
Posts: 155
Joined: Tue Mar 20, 2012 12:55 am

### Re: Putnam, 1962: Summing Binomials

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, $\displaystyle\binom{n}{r}=\frac{n}{r}\binom{n-1}{r-1}$ "Rhythm is mathematics of the sub-conscious."
Some-Angle Related Problems;

SANZEED
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Joined: Wed Dec 28, 2011 6:45 pm

### Re: Putnam, 1962: Summing Binomials Sorry, I have edited the typo.
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sm.joty
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Location: Dhaka

### 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.........

SANZEED
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Joined: Wed Dec 28, 2011 6:45 pm

### Re: Putnam, 1962: Summing Binomials

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.
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Tahmid Hasan
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Joined: Thu Dec 09, 2010 5:34 pm

### Re: Putnam, 1962: Summing Binomials

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 $2^{n-2}(2n+n^2-n)=n(n+1)2^{n-2}$.
বড় ভালবাসি তোমায়,মা

Masum
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### 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 $0$.

Masum
Posts: 592
Joined: Tue Dec 07, 2010 1:12 pm
Now, find the sum $\sum_{i=0}^n\binom nii^3$
The same trick should work. One one thing is neutral in the universe, that is $0$.