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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

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HINT:

$$k\cdot\binom nk=k\cdot\frac{n!}{(n-k)!\cdot k!}=k\cdot n\frac{(n-1)!}{\{(n-1)-(k-1)\}!\cdot (k-1)!\cdot k}=n\binom{n-1}{k-1}$$

Now, use the above method to find $\displaystyle k(k-1)\binom nk=n(n-1)\binom{n-2}{k-2}$

Again $\displaystyle k^2=k(k-1)+k,$

In general $$\sum_{0\le r\le m}a_rk^m=b_0k(k-1)\cdot\{k-(m-1)\}+b_1k(k-1)\cdot\{k-(m-2)\}+\cdots+b_m$$

where $a_r,0\le r\le m$ are given constants and $b_s,0\le s\le m$ are arbitrary constants to be determined by comparing the coefficients of the different powers of $k$

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  • $\begingroup$ I like your answer better than mine as it uses binomial identities. $\endgroup$ – user44197 Dec 31 '13 at 6:26
  • $\begingroup$ @user44197, thanks for your feedback. But, I could not solve the second Question of math.stackexchange.com/questions/437523/… w/o calculus $\endgroup$ – lab bhattacharjee Dec 31 '13 at 6:30
  • $\begingroup$ Thanks a lot! That was answer I was looking for. Stupidly, wasn't able to extend it to the quadratic form. $\endgroup$ – Avaneesh Narla Jan 4 '14 at 18:49
  • $\begingroup$ @AvaneeshNarla, nice to hear the 2nd statement $\endgroup$ – lab bhattacharjee Jan 4 '14 at 18:52
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$$ (p+q)^n = \sum \binom{n}{k} p^k q^{n-k}$$

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 $$ \sum k^2 \binom{n}{k} p^{k} q^{n-k} = n(n-1) p^2 +np$$

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  • $\begingroup$ I forgot to mention $q=1-p$. $\endgroup$ – user44197 Dec 31 '13 at 6:50
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I hope this helps. Let $1-p=q.$

$$\sum_{k=1}^{n} k\cdot \binom{n}{k}p^k\cdot q^{n-k}=\sum_{k=1}^{n} k・n!/{k!(n-k)!}\cdot p^k\cdot q^{n-k}$$ $$=\sum_{k=1}^{n}n!/{(k-1)!(n-k)!}\cdot p^k\cdot q^{n-k}$$ $$=n\sum_{k=1}^{n}(n-1)!/{(k-1)!(n-k)!}\cdot p^k\cdot p^{n-k}$$ $$=np\sum_{k=1}^{n}\binom{n-1}{k-1}\cdot p^{k-1}\cdot q^{n-k}=np(p+q)^{n-1}=np$$ On the other hand,

$$\sum_{k=1}^{n} k^2\cdot\binom{n}{k}\cdot p^k\cdot q^{n-k}=\sum_{k=1}^{n} k^2\cdot n!/{k!(n-k)!}\cdot p^k\cdot q^{n-k}$$ $$=\sum_{k=1}^{n} {k(k-1)+k}・n!/{k!(n-k)!}\cdot p^k\cdot q^{n-k}$$ $$=\sum_{k=1}^{n} k(k-1)\cdot n!/{k!(n-k)!}\cdot p^k・q^{n-k}+\sum_{k=1}^{n} k\cdot n!/{k!(n-k)!}\cdot p^k\cdot q^{n-k}$$ $$=\sum_{k=1}^{n}n!/{(k-2)!(n-k)!}\cdot p^k\cdot q^{n-k}+\sum_{k=1}^{n} n!/{(k-1)!(n-k)!}・p^k・q^{n-k}$$ $$=n(n-1)\sum_{k=1}^{n}(n-2)!/{(k-2)!(n-k)!}・p^k・q^{n-k}$$$$+n\sum_{k=1}^{n} (n-1)!/{(k-1)!(n-k)!}・p^k・q^{n-k}$$ $$=n(n-1)\sum_{k=1}^{n}\binom{n-2}{k-2}・p^k・q^{n-k}+n\sum_{k=1}^{n}\binom{n-1}{k-1} ・p^k・q^{n-k}$$ $$=n(n-1)p^2\sum_{k=1}^{n}\binom{n-2}{k-2}・p^{k-2}・q^{n-k}+np\sum_{k=1}^{n} \binom{n-1}{k-1}・p^{k-1}・q^{n-k}$$ $$ =n(n-1)p^2(p+q)^{n-2}+np(p+q)^{n-1}=n(n-1)p^2+np$$

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  • $\begingroup$ Thanks a lot! Though lab bhattacharjee did the same thing in summary. I referenced a version of your elongated solution, I hope you are fine with that. $\endgroup$ – Avaneesh Narla Jan 4 '14 at 18:48

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