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

Question

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

Is it possible to use $\sum\limits_{k=0}^{n}{n\choose k}(\frac{1}{3})^k(\frac{2}{3})^{n-k}=1$ (by binomial theorem) and $\sum\limits_{k=0}^{n}k^2=\frac{n(n+1)(2n+1)}{6}$?

Could someone give a hint on this problem?

Answer 1

If you know the gamma function:

$$\sum_{k=0}^{n}k^2{n\choose k}\left(\frac{1}{3}\right)^k\left(\frac{2}{3}\right)^{n-k}=\sum_{k=0}^{n}k^2 3^{-n}2^{n-k}{n\choose k}=$$ $$\sum_{k=0}^{n}\frac{k\cdot 3^{-n}2^{n-k}n!}{\Gamma(k)(n-k)!}=\sum_{k=0}^{n}\frac{k\cdot 2^{n-k}n!}{3^n\Gamma(k)(n-k)!}=\frac{n}{9}\left(n+2\right)$$

Answer 2

HINT

The pdf binomial distribution with parameter $p$ is given by:

$${n \choose k}p^k(1-p)^{n-k}$$

What is $p$ in this case? This problem is easier if you know the variance of this distribution

Answer 3

Your sum is $$S_n=(\frac{2}{3})^n\sum_{k=0}^{n}k^2{n\choose k}(\frac{1}{2})^k$$

Now if $\displaystyle \tau=x\frac{d}{dx}$, we have $\displaystyle\tau(x^k)=kx^k$, $\displaystyle\tau^2(x^k)=k^2x^k$. As $\displaystyle (1+x)^n=\sum_{k=0}^n {n\choose k}x^k$, we get $$\sum_{k=0}^n k^2{n \choose k} x^k=\tau^2 ((1+x)^n)=nx(1+x)^{n-1}+n(n-1)x^2(1+x)^{n-2}$$ and putting $\displaystyle x=\frac{1}{2}$, it is easy to finish.