Evaluating $\sum_{r \in \mathbb{N}} (n-2r+1)^2 \binom{n}{2r-1}$

Question

I am seeking to evaluate the sum $$S=\sum_{r \in \mathbb{N}} (n-2r+1)^2 \binom{n}{2r-1} \\ =(n-1)^2 \binom{n}{1}+(n-3)^2 \binom{n}{3}+(n-5)^2\binom{n}{5}+\cdots$$

I re-wrote the sum as $$S=(n+1)^2\sum_{r \in \mathbb{N}} \binom{n}{2r-1}-4(n+1)\sum_{r\in\mathbb{N}}r\binom{n}{2r-1}+4\sum_{r\in\mathbb{N}}r^2\binom{n}{2r-1}$$

It is clear from the binomial expansion of $(1+x)^n$, (by plugging $x=\pm1$ ) we get the value of the sum $\displaystyle \sum_{r\in\mathbb{N}}\binom{n}{2r-1}$ as $2^{n-1}$.

Now, after this I can find the value of the sums $\displaystyle \sum_{r\in\mathbb{N}}r\binom{n}{2r-1}$ and $\displaystyle \sum_{r\in\mathbb{N}}r^2\binom{n}{2r-1}$ by considering the binomial expansion of $\displaystyle \dfrac{(1+x)^n-(1-x)^n}{2}$ and substituting $x$ for $\sqrt{x}$ and then differentiating, and by manipulations, I will be able to find the desired sums.

However, this method gets very lengthy, especially when doing differentiation for finding $\displaystyle \sum_{r\in\mathbb{N}}r^2\binom{n}{2r-1}$.

I am hoping to see some other different approaches which don't involve so much calculation and are easy to understand.

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The final closed form is $n(n+1)2^{n-3}$.

Answer 1

We can rewrite the sum as:

$$\begin{aligned}S&=\sum_{r \in \mathbb{N}} (n-2r+1)^2 {n \choose 2r-1} \\&= \sum_{r \in \mathbb{N}} (n-2r+1)^2 {n \choose n-2r+1} \\&= \sum_{r \in \mathbb{N}} (n-2r+1)^2 \frac{n!}{(n-2r+1)!(2r-1)!} \\&= \sum_{r \in \mathbb{N}} n(n-2r+1) \frac{(n-1)!}{(n-2r)!(2r-1)!} \\&= \sum_{r \in \mathbb{N}} n(n-2r+1) {n-1 \choose 2r-1} \\&= n^2 \sum_{r \in \mathbb{N}} {n-1 \choose 2r-1} - n \sum_{r \in \mathbb{N}} (2r-1){n-1 \choose 2r-1} \\&= n^2 \frac{2^{n-1}}{2} - n(n-1) \frac{2^{n-2}}{2} \\&= n2^{n-3}(2n-n+1) \\&= n(n+1)2^{n-3}\end{aligned}$$

Answer 2

For a combinatorial proof, consider choosing from $n$ people a committee with an odd number of regular members and a secretary and treasurer (which might be the same person). The LHS conditions on the number $2r-1$ of regular members. The RHS first selects the secretary and treasurer and then selects an odd number of remaining members. If one person is both secretary and treasurer, then half of the subsets of the remaining $n-1$ people are odd. If the secretary and treasurer are different people, then half of the subsets of the remaining $n-2$ people are odd. The sum of the two counts is: $$n\cdot\frac{2^{n-1}}{2} + n(n-1) \frac{2^{n-2}}{2} = (2n+(n^2-n))2^{n-3} = n(n+1)2^{n-3}$$

Answer 3

Hint:

Let

$$(n-2r+1)^2=a+b(2r-1)+c(2r-1)(2r-2)$$

Set $2r-1=0\implies a=n^2$

$2r-2=0\implies a+b=(n-1)^2\implies b=?$

Compare the coefficients of $r^2,c=1$

Now $$(2r-1)(2r-2)\cdot\binom n{2r-1}=n(n-1)\binom{n-2}{2r-3}$$

Answer 4

Use the chain rule.

$$ f(x)=\displaystyle \dfrac{(1+x)^n-(1-x)^n}{2}$$

We want $\frac{d^2 f(\sqrt{x})}{dx^2}$ which can be simplified as:

$$ \frac{d^2 f(\sqrt{x} )}{dx^2} = \frac12 \frac{d}{dx}( \frac{df}{dx}|_{\sqrt{x} } \frac{1}{ \sqrt{x}}) =\frac12 \left[ \frac{d^2f}{dx^2}|_{\sqrt{x}} \frac{1}{2x} -\frac{1}{2} \frac{df}{dx}|_{\sqrt{x}} \frac{1}{x^{\frac32}}\right] $$

Now, here's the interesting part put $x=1$

$$ \frac{d^2 f \circ (\sqrt{x})}{dx^2}|_{x=1}= \frac{1}{4} \left[\frac{d^2 f}{dx^2}|_{1} - \frac{df}{dx}|_{1} \right]$$

And, the right side is now easy to evaluate on the given expression