Sum of Binomial Coefficients Times a Polynomial

Question

Is there a closed for expression for,

$\displaystyle\sum_{k=0}^n {n\choose k}k^2$

It holds that,

$\displaystyle\sum_{k=0}^n {n\choose k}k=n2^{n-1}$

Is there a generalization for higher degrees?

Answer 1

Binomial Coefficients as Polynomials

Note that since $$ \binom{k}{j}=\frac{k(k-1)\cdots(k-j+1)}{j!}\tag{1} $$ $\binom{k}{j}$ can be viewed as a degree $j$ polynomial in $k$. We can write any degree $j$ polynomial in $k$ as a linear combination of $\binom{k}{0}$, $\binom{k}{1}$, $\dots$, $\binom{k}{j}$. For example, $$ \begin{align} 1&=\binom{k}{0}\\ k&=\binom{k}{1}\\ k^2&=2\binom{k}{2}+\binom{k}{1}\\ \end{align}\tag{2} $$

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Multiplying a Binomial Coefficient $$ \begin{align} k\binom{k}{j} &=\left[(j+1)\frac{k+1}{j+1}-1\right]\binom{k}{j}\tag{3a}\\ &=(j+1)\binom{k+1}{j+1}-\binom{k}{j}\tag{3b}\\ &=(j+1)\binom{k}{j+1}+(j+1)\binom{k}{j}-\binom{k}{j}\tag{3c}\\ &=(j+1)\binom{k}{j+1}+j\binom{k}{j}\tag{3d} \end{align} $$ Explanation:
$(3a)$: creative rewriting of $k$
$(3b)$: $\binom{k+1}{j+1}=\frac{k+1}{j+1}\binom{k}{j}$
$(3c)$: $\binom{k+1}{j+1}=\binom{k}{j+1}+\binom{k}{j}$
$(3d)$: algebra

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Applying $\boldsymbol{(3)}$

We can recursively extend the list started in $(2)$: $$ \begin{align} k^3 &=k\cdot k^2\\ &=2k\binom{k}{2}+k\binom{k}{1}\\ &=2\left(3\binom{k}{3}+2\binom{k}{2}\right)+\left(2\binom{k}{2}+\binom{k}{1}\right)\\ &=6\binom{k}{3}+6\binom{k}{2}+\binom{k}{1}\tag{4} \end{align} $$ $$ \begin{align} k^4 &=k\cdot k^3\\ &=6k\binom{k}{3}+6k\binom{k}{2}+k\binom{k}{1}\\ &=6\left(4\binom{k}{4}+3\binom{k}{3}\right)+6\left(3\binom{k}{3}+2\binom{k}{2}\right)+\left(2\binom{k}{2}+\binom{k}{1}\right)\\ &=24\binom{k}{4}+36\binom{k}{3}+14\binom{k}{2}+\binom{k}{1}\tag{5} \end{align} $$

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How This Helps

Once we have written the polynomial as a linear combination of binomial coefficients, we can use the following formula to simplify the final sums: $$ \begin{align} \sum_{k=0}^n\binom{n}{k}\binom{k}{j} &=\sum_{k=0}^n\binom{n}{j}\binom{n-j}{k-j}\\ &=\binom{n}{j}2^{n-j}\tag{6} \end{align} $$

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Application to the Question

We can apply the formulas derived above to get the following generalizations

$$ \begin{align} \sum_{k=0}^n\binom{n}{k}k^2 &=\sum_{k=0}^n\binom{n}{k}\left(2\binom{k}{2}+\binom{k}{1}\right)\\ &=2\binom{n}{2}2^{n-2}+\binom{n}{1}2^{n-1}\\ &=\bbox[5px,border:2px solid #C0A000]{2^{n-1}\left(\binom{n}{2}+\binom{n}{1}\right)}\tag{7} \end{align} $$ $$ \begin{align} \sum_{k=0}^n\binom{n}{k}k^3 &=\sum_{k=0}^n\binom{n}{k}\left(6\binom{k}{3}+6\binom{k}{2}+\binom{k}{1}\right)\\ &=6\binom{n}{3}2^{n-3}+6\binom{n}{2}2^{n-2}+\binom{n}{1}2^{n-1}\\ &=\bbox[5px,border:2px solid #C0A000]{2^{n-2}\left(3\binom{n}{3}+6\binom{n}{2}+2\binom{n}{1}\right)}\tag{8} \end{align} $$ $$ \begin{align} \sum_{k=0}^n\binom{n}{k}k^4 &=\sum_{k=0}^n\binom{n}{k}\left(24\binom{k}{4}+36\binom{k}{3}+14\binom{k}{2}+\binom{k}{1}\right)\\ &=24\binom{n}{4}2^{n-4}+36\binom{n}{3}2^{n-3}+14\binom{n}{2}2^{n-2}+\binom{n}{1}2^{n-1}\\ &=\bbox[5px,border:2px solid #C0A000]{2^{n-1}\left(3\binom{n}{4}+9\binom{n}{3}+7\binom{n}{2}+\binom{n}{1}\right)}\tag{9} \end{align} $$

The formulas above can be extended to give formulas for as high a degree polynomial as one wishes.

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Using Stirling Numbers

Stirling Numbers of the Second Kind allows us to easily write monomials as linear combinations of binomial coefficients with the following formula $$ \newcommand{\stirtwo}[2]{\left\{{#1}\atop{#2}\right\}} k^j=\sum_{i=0}^ji!\stirtwo{j}{i}\binom{k}{i}\tag{10} $$ Summing, we get $$ \begin{align} \sum_{k=0}^n\binom{n}{k}k^jx^k &=\sum_{k=0}^n\binom{n}{k}\sum_{i=0}^ji!\stirtwo{j}{i}\binom{k}{i}x^k\\ &=\sum_{i=0}^ji!\stirtwo{j}{i}\sum_{k=0}^n\binom{n}{k}\binom{k}{i}x^k\\ &=\sum_{i=0}^ji!\stirtwo{j}{i}\sum_{k=0}^n\binom{n}{i}\binom{n-i}{k-i}x^k\\ &=\bbox[5px,border:2px solid #C0A000]{(1+x)^n\sum_{i=0}^ji!\stirtwo{j}{i}\binom{n}{i}\left(\frac x{1+x}\right)^i}\tag{11} \end{align} $$ Formulas $(7)$-$(9)$ can be derived from $(11)$ by setting $x=1$.

Answer 2

For a different solution for a general case that the given by @JackD'Aurizio you want to transform the polynomial $p(k)$ in the expression

$$\sum_k\binom{n}{k}p(k)$$

in a polynomial based in falling factorials. By example

$$\sum_k\binom{n}{k}k^2=\sum_k\binom{n}{k}(k^{\underline{2}}+k)$$

Then the binomial sum can be simplified extracting falling factorials of $n$ out of the summation and (probably) changing the index of the sums.

To pass from $p(k)$ to a "polynomial" of falling factorials you will need to know that

$$k^n=\sum_j \left\{{ n\atop j }\right\}k^{\underline j}$$

for $n\ge 0$, where the symbols $\left\{{ n\atop j }\right\}$ are Stirling numbers of the second kind.

This link have an easy algorithm to pass an entire polynomial to a "polynomial" of falling factorials.

Answer 3

For the factor $k$,

$$\sum_{k=1}^n {n\choose k}k=\sum_{k=1}^n \frac{n!}{(n-k)!k!}k =n\sum_{k=1}^n \frac{(n-1)!}{(n-k)!(k-1)!}\\ =n\sum_{k=0}^{n-1} {n\choose k}=n2^{n-1}.$$

Similarly for $(k-1)k$,

$$\sum_{k=2}^n {n\choose k}(k-1)k=\sum_{k=1}^n \frac{n!}{(n-k)!k!}(k-1)k=(n-1)n\sum_{k=1}^n \frac{(n-2)!}{(n-k)!(k-2)!}\\ =(n-1)n\sum_{k=0}^{n-2} {n\choose k}=(n-1)n2^{n-2}.$$

More generally,

$$\sum_{k=m}^n {n\choose k}(k-m+1)\cdots(k-1)k\\ =\sum_{k=1}^n \frac{n!}{(n-k)!k!}(k-m+1)\cdots(k-1)k\\ =(n-m+1)\cdots(n-1)n\sum_{k=1}^n \frac{(n-m)!}{(n-k)!(k-m)!}\\ =(n-1)n\sum_{k=0}^{n-m} {n\choose k}=(n-m+1)(n-1)n2^{n-m}.$$

Then,

$$k^2=(k-1)k+k,\\ k^3=(k-2)(k-1)k+3k^2-2k=(k-2)(k-1)k+3(k-1)k+k\\ \cdots$$ and you can convert any polynomial to a sum of falling factorials.

Notice that the $m$ first terms in the summations must be added separately.