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I am trying to generalise the closed form expression for $k$ of the following sum using diffferentiation techniques.

$$\sum_{r=0}^{n}r^{k}{n\choose r}$$

We know the following results through trivial application of differentiation followed by multiplication at each step, and at the final step plugging in $x=1$.

$$\begin{aligned}\sum_{r=0}^{n}r{n\choose r}&=n2^{n-1}\\ \sum_{r=0}^{n}r^2{n\choose r}&=n(n+1)2^{n-2}\\\sum_{r=0}^{n}r^{3}{n\choose r}&=n^{2}(n+3)2^{n-3}\end{aligned}$$

Clearly there seems to be some patterns, as for the exponent, the degree of the polynomial function. I am not able to extend a similar pattern for $k=4$. Any ideas are appreciated. Thanks.

Paras Khosla
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2 Answers2

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For every nonnegative integer $k$, and every integer $m\geq k$, you have $\frac{d^k}{dx^k}(1+x)^m=\frac{m!}{(m-k)!}(1+x)^{m-k}$ and \begin{eqnarray*} \frac{d^k}{dx^k}(1+x)^m&=& \frac{d^k}{dx^k}\sum_{s=0}^m\binom{m}{s}x^s\\ &=&\sum_{s=0}^m\binom{m}{s}\frac{d^k}{dx^k}x^s\\ &=&\sum_{s=k}^m\binom{m}{s}\frac{s!}{(s-k)!}x^{s-k}\\ &=&\sum_{r=0}^n\binom{n}{r}\frac{(r+k)!}{r!}x^r\\ \end{eqnarray*} Now note that $$\frac{(r+k)!}{r!}=\prod_{i=0}^{k-1}(r+i)=r^k+\sum_{j=1}^ks_jr^{k-j},$$ where $s_j$ denotes the $j$-th symmetric polynomial in $k$ variables evaluated at $\{1,2,\ldots,k\}$. It follows that \begin{eqnarray*} \sum_{r=0}^nr^k\binom{n}{r}x^r&=&\frac{(n+k)!}{n!}(1+x)^n-\sum_{r=0}^n\binom{n}{r}\left(\sum_{j=0}^{k-1}s_jr^{k-j}\right)x^r\\ &=&\frac{(n+k)!}{n!}(1+x)^n-\sum_{j=1}^ks_j\sum_{r=0}^nr^{k-j}\binom{n}{r}x^r, \end{eqnarray*} and so plugging in $x=1$ and setting $S(n,k):=\sum_{r=0}^nr^k\binom{n}{r}$ shows that $$S(n,k)=\sum_{r=0}^nr^k\binom{n}{r}=\frac{(n+k)!}{n!}2^n-\sum_{j=1}^ks_jS(n,k-j).$$

Servaes
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Let $ n\geq 2 $.

It is possible to prove that $ \mathcal{B}=\left(1,X,\cdots, X\left(X-1\right)\cdots\left(X-n+1\right)\right) $ is a basis of $ \mathbb{R}_{n}\left[X\right] $. To do so, we just need to prove the linear independence of $ 1 $, $ X $, $\cdots $, $X-n+1 $.

Anyway, that means there exists an unique $ \left(n+1\right)-$tuple $ \left(\alpha_{n,k}\right)_{0\leq k\leq n} $ such that : $$ X^{n}=\sum_{k=0}^{n}{\alpha_{n,k}X\cdots \left(X-k+1\right)} $$

Therefore if $ n,k\in\mathbb{N} $, we have for all $ i\in\mathbb{N} $ : $$ i^{k}=\sum_{j=0}^{k}{\alpha_{k,j}i\cdots\left(i-j+1\right)} $$

Thus : \begin{aligned}\sum_{i=0}^{n}{i^{k}\binom{n}{i}}&=\sum_{i=0}^{n}{\sum_{j=0}^{k}{\alpha_{k,j}i\cdots\left(i-j+1\right)\binom{n}{i}}}\\ &=\sum_{j=0}^{k}{\alpha_{k,j}\sum_{i=j}^{n}{i\cdots\left(i-j+1\right)\binom{n}{i}}}\\ &=\sum_{j=0}^{k}{\alpha_{k,j}\sum_{i=j}^{n}{n\cdots\left(n-j+1\right)\binom{n-j}{i-j}}}\\ \sum_{i=0}^{n}{i^{k}\binom{n}{i}}&=\sum_{j=0}^{k}{\alpha_{k,j}n\cdots\left(n-j+1\right)2^{n-j}}\end{aligned}

The $ \left\lbrace\alpha_{k,j}\right\rbrace_{0\leq j\leq k} $ are easy to find in particular cases, calculating the sum should be easy using the previous result, after finding the $ \alpha_{k,j} $.

Their might be a recurrence relation between the $ \alpha_{k,j} $. Finding it will help us to generalise further.

CHAMSI
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