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



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

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 $k$

  • $\begingroup$ 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

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

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

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

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,

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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.