Evaluating sum of a combination series

Question

Please help me evaluate the following series.

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

Answer 1

Method 1 (Generating function oriented)

$$ \begin{align} \sum_{k=0}^n k \binom{n}{k}^2 \stackrel{\color{blue}{^{[1]}}}{=} & \sum_{k=0}^n\sum_{l=0}^n k\binom{n}{k}\binom{n}{l}\delta_{kl}\\ \stackrel{\color{blue}{^{[2]}}}{=} & \sum_{k=0}^n\sum_{l=0}^n k\binom{n}{k}\binom{n}{l}\left(\int_0^{2\pi}e^{i(k-l)\theta}\frac{d\theta}{2\pi}\right)\\ \stackrel{\color{blue}{^{[3]}}}{=} & \lim_{s\to 1} s\frac{d}{ds}\left[\sum_{k=0}^n\sum_{l=0}^n \binom{n}{k}\binom{n}{l}\left(s^k\int_0^{2\pi}e^{i(k-l)\theta}\frac{d\theta}{2\pi}\right)\right]\\ = & \lim_{s\to 1} s\frac{d}{ds}\left[\int_0^{2\pi} ( 1 + s e^{i\theta})^n( 1 + e^{-i\theta})^n\frac{d\theta}{2\pi}\right]\\ = & n \int_0^{2\pi} e^{i\theta}( 1 + e^{i\theta})^{n-1}( 1 + e^{-i\theta})^n\frac{d\theta}{2\pi}\\ = & n \int_0^{2\pi} e^{-i(n-1)\theta} ( 1 + e^{i\theta} )^{2n-1} \frac{d\theta}{2\pi}\\ \stackrel{\color{blue}{^{[4]}}}{=} & n \binom{2n-1}{n-1} \end{align}$$

Notes

• $\color{blue}{[1]}$ $\delta_{kl}$ is the Kronecker delta.

• $\color{blue}{[2]}$ Given two sequences $a_k$, $b_k$ and corresponding generating functions $$a(z) = \sum_{k} a_k z^k\quad\text{ and }\quad b(z) = \sum_{k} b_k z^k.$$ the introduction of a Kronecker delta followed by a integral representation: $$\delta_{kl} = \int_0^{2\pi} e^{i(k-l)\theta} \frac{d\theta}{2\pi}$$ is one trick to obtain a generating function for the product sequence $a_kb_k$: $$\sum_{k}a_kb_k z^k = \int_{0}^{2\pi} a(\sqrt{z}e^{i\theta}) b(\sqrt{z} e^{-i\theta}) \frac{d\theta}{2\pi}$$

• $\color{blue}{[3]}$ The introduction of an auxiliary variable, $s$ in this case, and then replace $k$ by $s\frac{d}{ds} s^k$ is another common trick to deal with combinatorial sums that involve binomial coefficients.
• $\color{blue}{[4]}$ Let $z = e^{i\theta}$. The $e^{-i(n-1)\theta}$ factor in the integrand pick up and only pick up the coefficient of $z^n$ in the polynomial $(1+z)^{2n-1}$. We then apply binomial theorem to $(1+z)^{2n-1}$ to obtain the last expression.

Method 2 (Combinatorial)

Notice $\displaystyle\quad\binom{n}{k} = \binom{n}{n-k}\quad$, we have:

$$\sum_{k=0}^n k \binom{n}{k}^2 = \sum_{k=0}^n (n-k)\binom{n}{n-k}^2 = \sum_{k=0}^n(n-k)\binom{n}{k}^2$$

This implies $$\sum_{k=0}^n k \binom{n}{k}^2 = \frac{n}{2}\sum_{k=0}^n \binom{n}{k}^2 = \frac{n}{2}\sum_{k=0}^n \binom{n}{k} \binom{n}{n-k}$$ The last sum in R.H.S has a combinatorial realization.

If you are given $2n$ balls and asked to pick $n$ balls out of them. One way to count the number of possibilities is first split the $2n$ balls into two groups of $n$ balls. You then pick $0 \le k \le n$ from the first group and then $n-k$ balls from the second group. For each $k$, the number of possibilities is $\displaystyle \binom{n}{k} \binom{n}{n-k}$. Since we know the sum of these numbers will give us the total number of possibilities $\displaystyle \binom{2n}{n}$, we obtain following identity:

$$\sum_{k=0}^n \binom{n}{k} \binom{n}{n-k} = \binom{2n}{n}$$

Using this, we can conclude

$$\sum_{k=0}^n k \binom{n}{k}^2 = \frac{n}{2}\binom{2n}{n} = n\binom{2n-1}{n-1}$$

Answer 2

I take it you mean the sum is over $k$.

You can start by writing $k\binom{n}{k} = n\binom{n-1}{n-k}$. This effectively reduces the problem to evaluating $\sum_k \binom{n-1}{n-k}\binom{n}{k}$. Then you can apply a Vandermonde convolution identity.