Could someone help me find:
$$\sum_{k}k \binom{n}{k}p^k(1-p)^{n-k}\\
and \sum_{k}k^2 \binom{n}{k}p^k(1-p)^{n-k}\\
0\leq p\leq 1, k\in N, n\ggg k $$
I know the answer to the first one is np, and the second is np(np-p+1) by simulation. But I am not able to prove them.
Can you generalise for all powers of k? It is obvious for 0.
$\sum_{k} \binom{n}{k}p^k(1-p)^{n-k}=[p+(1-p)]^n=1$
Wikipedia has a few nice solutions for similar series.
http://en.wikipedia.org/wiki/Binomial_coefficient#Series_involving_binomial_coefficients
Best Answer
$$ (p+q)^n = \sum \binom{n}{k} p^k q^{n-k}$$
differentiate with respect to $p$ to get $$ n (p+q)^{n-1} =\sum k \binom{n}{k} p^{k-1} q^{n-k} = \frac{1}{p} \sum k \binom{n}{k} p^{k} q^{n-k} \tag 1$$
Differentiate once more $$ n (n-1)(p+q)^{n-2} =\sum k (k-1) \binom{n}{k} p^{k-2} q^{n-k} = \frac{1}{p^2} \sum k (k-1)\binom{n}{k} p^{k} q^{n-k} \tag 2$$
from (1) you get $$ \sum k \binom{n}{k} p^{k} q^{n-k} = np$$ from (2) you get $$ \sum k^2 \binom{n}{k} p^{k} q^{n-k} = n(n-1) p^2 +np$$