Evaluating Finite Sum: Homework Statement

[gruba]

Homework Statement

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

Homework Equations

-Binomial theorem

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)?

 

[jk22]

How do you compute the exponent of 2/3 ?

 

[gruba]

How do you compute the exponent of 2/3 ?

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.

 

[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.

 

[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.

 

[HallsofIvy]

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.

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.