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I have the following question: I am trying to prove that $\sum_{k=0}^{n} {\binom{N}{k}}^2 = \binom{2N}{k}$.

I tried proving by induction $\sum_{k=0}^{n} {\binom{N}{k}}^2$,

so I substituted k = 0 , 1,2,3,4..n, and I got $\sum_{k=0}^{n} {\binom{N}{k}}^2= 1 + n^2 + 0.25(n^4-2n^3+n^2)+..+(1/n!)^2$ now I dunno how to move farther. Should I be doing the same with the other term to prove them equal

Phicar
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sera
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  • Hi, welcome. I edited your question, you have to use the $ sign to add latex. You can edit your question to check how and correct the rest of the question. – Phicar Feb 22 '21 at 19:03
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    What do you mean by $k$ in the RHS? $k$ is the integer "variable" of the LHS it isn't a fixed value – Aryan Feb 22 '21 at 19:05
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    Also, probably that should be $n$ not $N$ – Aryan Feb 22 '21 at 19:06
  • my bad. you are right: – sera Feb 22 '21 at 19:10
  • @Aryaaaaan my bad you are right. it should be like that . $\sum_{k=0}^{n} {\binom{n}{k}}^2 = \binom{2n}{n}$ – sera Feb 22 '21 at 19:13
  • Are you familiar with the notion of combinatorial proofs? Rewrite the LHS as the sum of ${n\choose k}{n\choose n-k}$ instead. The RHS counts the number of ways to choose $k$ elements from a set of $2n$ elements. Think of this set as being made of two halves, each containing $n$ elements. Can you see how the LHS counts the same thing? – Kevin Long Feb 22 '21 at 19:57

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