# Prove <munderover> &#x2211;<!-- ∑ --> <mrow class="MJX-TeXAtom-ORD"> k =

Prove $\sum _{k=0}^{n}\frac{\left(2k\right)!\left(2n-2k\right)!}{{2}^{2n}\left(2k+1\right)\left[k!\left(n-k\right)!{\right]}^{2}}=\frac{\left[{2}^{n}n!{\right]}^{2}}{\left(2n+1\right)!}$ using mathematical induction
Since I already know the way using combinatorial proof, I am trying to prove it by mathematical induction but it is not easy to show.
I derived that
${S}_{n}=\frac{2n}{2n+1}{S}_{n-1}$ from RHS, but how can I proceed from this? I mean, how can I organize the LHS? I cannot organize LHS properly because of different .
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Step 1
First of all notice that:
$\left(2n+1\right)!=1\cdot 2\cdot 3\cdot 4\cdot \dots \cdot 2n\cdot \left(2n+1\right)=\left[1\cdot 3\cdot 5\cdot \dots \cdot \left(2n+1\right)\right]\cdot {2}^{n}\left[1\cdot 2\cdot \dots \cdot n\right]=\left(2n+1\right)!!\cdot {2}^{n}\cdot n!$
Step 2
Based on the first hint, you want to prove that:
${S}_{n}=\frac{\left[{2}^{n}n!{\right]}^{2}}{\left(2n+1\right)!}=\frac{{2}^{n}n!}{\left(2n+1\right)!!}$
Step 3
What does the second hint and the rule:
${S}_{n}=\frac{2n}{2n+1}{S}_{n-1}$ have in common?