sum of multiple of two binomial coefficient

Question

I'm trying to show this equality. I try to expand it but I have no idea to go on.Thanks in advance for your help. The equality is

$$\sum_{i=0}^{n-1}(2i+1){n-2\choose n-1-i}{n\choose i+1}=2(n-1){2n-3\choose n}+{2(n-1)\choose n}+2{2n-3\choose n-1}$$

I write the left side of equality as below

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

I appreciate help me to continue.

Answer 1

Hints:

1. Rewrite $2i+1$ as $2(i+1)-1$, and split into two sums.
2. For the left sum, apply $(i+1)\binom{n}{i+1}=n\binom{n-1}{i}$.
3. For both sums, use Vandermonde.

Answer 2

Let $j=i+1$ you can write your sum as: $$\sum_{j=1}^n(2j-1)\binom{n-2}{n-j}\binom{n}{j}=2\left[\sum_{j=1}^nj\binom{n-2}{n-j}\binom{n}{j}\right]-\left[\sum_{j=1}^n\binom{n-2}{n-j}\binom{n}{j}\right]$$ You can use the hypergeometrics sums: $$\sum_{j=\max\{0,d+m-N\}}^{\min\{m,d\}}\binom{d}{j}\binom{N-d}{m-j}=\binom{N}{m},\qquad\sum_{j=\max\{0,d+m-N\}}^{\min\{m,d\}}j\binom{d}{j}\binom{N-d}{m-j}=\binom{N}{m}\frac{md}{N}$$ You can choose $d=n$, $m=n$ and $N=2n-2$ to prove: \begin{align} \sum_{j=1}^n\binom{n}{j}\binom{n-2}{n-j}&=\sum_{j=2}^n\binom{n}{j}\binom{n-2}{n-j}+n\binom{n-2}{n-1}=\binom{2n-2}{n}+n\binom{n-2}{n-1}\\ \sum_{j=1}^nj\binom{n}{j}\binom{n-2}{n-j}&=\sum_{j=2}^nj\binom{n}{j}\binom{n-2}{n-j}+n\binom{n-2}{n-1}=\binom{2n-2}{n}\frac{n^2}{2n-2}+n\binom{n-2}{n-1}\ \end{align} Finally \begin{align} \sum_{j=1}^n(2j-1)\binom{n-2}{n-j}\binom{n}{j}&=\binom{2n-2}{n}\left[2-\frac{n^2}{2n-2}\right]+n\binom{n-2}{n-1} \end{align} It is not what you were looking for :(