sum multiple of binomial cofficient

Question

I want to prove this equality : $$\sum_{i=0}^{n-1}2i {n-1\choose n-1-i}{n\choose i}=2(n-1){2(n-1)\choose n-1}$$ I write the left side of equality as follow but I don't know how I can continue. $\sum_{i=0}^{n-1}2i {n-1\choose n-1-i}{n\choose i}=2{n-1\choose n-2}{n\choose 1}+4{n-1\choose n-3}{n\choose 2}+6{n-1\choose n-4}{n\choose 3}+...+2(n-1){n-1\choose 0}{n\choose n-1}=2({n\choose 1}{n-1\choose n-2}+2{n\choose 2}{n-1\choose n-3}+3{n\choose 3}{n-1\choose n-4}+...+(n-1){n\choose n-1}{n-1\choose 0}=2(n{n-1\choose 0}{n-1\choose 1}+n{n-1\choose 1}{n-1\choose 2}+n{n-1\choose 2}{n-1\choose 3}+...+n{n-1\choose 1}{n-1\choose 0}=?$ I appreciate if anybody can help me. Thanks in advance.

Answer 1

$$2\left(n{n-1\choose 0}{n-1\choose 1}+n{n-1\choose 1}{n-1\choose 2}+n{n-1\choose 2}{n-1\choose 3}+...+n{n-1\choose 1}{n-1\choose 0}\right) \\=2n\sum_{k=0}^{n-2}\binom{n-1}{k}\binom{n-1}{n-2-k}=2n\binom{2n-2}{n-2}=2(n-1)\binom{2(n-1)}{n-1}$$