# Summation proof

1. Jan 13, 2008

### ehrenfest

1. The problem statement, all variables and given/known data
Does anyone know a clever way to prove that

$$\sum_{i=1}^{n}i^2 {n \choose i} = n(n+1) 2^{n-2}$$

where B(n,i) is n take i?

I can do it, but I had to divide into the cases of n = odd and n = even and it took about 1 page front and back. I'm sure there is a trick.

2. Relevant equations

3. The attempt at a solution

Last edited: Jan 13, 2008
2. Jan 13, 2008

### morphism

Take the second derivative of (1+x)^n and its binomial expansion, then mess with the index of the summation and set x=1. (I'm assuming your n+1 is actually n-1.)

3. Jan 13, 2008

### HallsofIvy

Staff Emeritus
??? It's obviously not true if you replace (n+1) by (n-1). Take n= 1. Then the lefthand side is 1. [iotex]n(n+1)2^{n-2}[/itex] becomes, for n= 1, $1(2)2^{-1}= 1$. If you replace (n+1) by (n-1), it becomes $$2(0)2^{-1}= 0[/itex]. 4. Jan 13, 2008 ### morphism You're right of course. I was a bit careless with my algebra. The identity I had in mind was: [tex]\sum_{i=0}^n i(i-1) {n \choose i} = n(n-1)2^{n-2}$$

But no worries - a similar trick can still be applied: Differentiate, multiply by x, and differentiate again!