Proof by induction

[Albi]

Guys, I'm trying to prove that $$\sum ^{n}_{k=1}k\binom{n}{k}^2= n\binom{2n-1}{n-1}$$And at the n+1th step I'm doing this $$\sum ^{n+1}_{k=1}k\binom{n}{k}^2=\sum ^{n}_{k=1}k\binom{n}{k}^2+ k\binom{n}{k}^2$$ and then I'm using the inductive hypothesis and I'm getting:$$n\binom{2n-1}{n-1}+$$ and I'm stuck right here I don't know how to write $k\binom{n}{k}^2$, can someone tell me what am I doing wrong.

 

[topsquark]

On the second line, the k in the second term on the RHS is n + 1. And I you also have to change the n in the upper "index" of the binomial coefficient when you go up one:
$$\sum_{k = 1}^{n + 1} k {n + 1 \choose k} ^2 = \sum_{k = 1}^{n} k {n + 1 \choose k} ^2 + (n + 1) {n + 1 \choose n + 1 } ^2$$
-Dan

 

[Albi]

On the second line, the k in the second term on the RHS is n + 1. And I you also have to change the n in the upper "index" of the binomial coefficient when you go up one:
$$\sum_{k = 1}^{n + 1} k {n + 1 \choose k} ^2 = \sum_{k = 1}^{n} k {n + 1 \choose k} ^2 + (n + 1) {n + 1 \choose n + 1 } ^2$$
-Dan

How do I use the inductive hypothesis then?

 

[topsquark]

How do I use the inductive hypothesis then?

Someone is going to have to be more clever than me.
$${n + 1 \choose k} = \dfrac{(n + 1)!}{k! (n + 1 - k)!} = \dfrac{(n + 1) n!}{k! (n + 1 - k) (n - k)!} = \dfrac{n + 1}{n + 1 - k} {n \choose k}$$
If we could get the k out of the coefficient on the RHS we could move it outside the summation. But I can't see any way to do that.

-Dan