# Evaluate finite sum

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1. Nov 26, 2015

### gruba

1. The problem statement, all variables and given/known data
Find $\sum\limits_{k=0}^{n}k^2{n\choose k}(\frac{1}{3})^k(\frac{2}{3})^{n-k}$

2. Relevant equations
-Binomial theorem

3. The attempt at a solution
I am using the binomial coefficient identity ${n\choose k}=\frac{n}{k}{{n-1}\choose {k-1}}$:

$$\sum\limits_{k=0}^{n}k^2{n\choose k}(\frac{1}{3})^k(\frac{2}{3})^{n-k}=\sum\limits_{k=1}^{n-1}\frac{n}{k}{{n-1}\choose {k-1}}(k-1)^2(\frac{1}{3})^{k-1}(\frac{2}{3})^{n-k+2}$$

What am I doing wrong here (sums are not equal)?

2. Nov 26, 2015

### jk22

How do you compute the exponent of 2/3 ?

3. Nov 26, 2015

### gruba

What do you mean? In the sum, $k$ is increased by $1$ and $n$ is decreased by $1$, so in the function $k$ is decreased and $n$ is increased.

4. Nov 26, 2015

### jk22

I think first we shall note the domain of validity of the binomial formula.

This imply we shall write a term outside the sum.

Are the k modified in the sum out side the binomial formula ?

I suppose the aim of using that formula two times is to get rid of the k squared.

5. Nov 26, 2015

### geoffrey159

In general, you have $(x+y)^n = \sum_{k = 0}^n \binom{n}{k} x^k y^{n-k}$

Take the derivative with respect to $x$ on each sign of the equation and multiply by $x$.
You get $nx(x+y)^{n-1} = \sum_{k = 0}^n k \binom{n}{k} x^k y^{n-k}$.

Now do it again and set $x$ and $y$ to one third and two thirds.

6. Nov 26, 2015

### HallsofIvy

Apparently you are trying to do this problem without knowing what $\sum_{k= 0}^n$ means.
n does NOT "decreas", n is fixed, the maximum value of k.

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