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I keep trying to prove this identity without being able to do it. I tried proving that : $\sum_{k=0}^{n}{2^{n-k}\binom{n+k}{k}}=\sum_{k=0}^{n}{2^{n}\binom{n}{k}}$ which we can observe by writing (using Newton's formula): $4^{n} = (2+2)^{n}=\sum_{k=0}^{n}{2^{n}\binom{n}{k}}$.

I thought that proving that : $2^{n-k}\binom{n+k}{k} = 2^{n}\binom{n}{k}$ but it turned out not to be the case, because it's a wrong equality.

Do you have any ideas how I can prove this ?

soufiane yes
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    This question is similar to: How to prove that $\sum_{i=0}^n 2^i\binom{2n-i}{n} = 4^n$.. If you believe it’s different, please edit the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. – Amrut Ayan Jul 24 '24 at 18:41
  • @whatamidoing Yes thank you. It's the same if we look at it as another formula of $4^{n}$, and also the approch is the same if we want to reason on this exact sum. But, I've found the solution to this exact problem in this question: https://math.stackexchange.com/questions/389099/proof-of-the-identity-2n-sum-limits-k-0n-2-k-binomnkk – soufiane yes Jul 24 '24 at 22:33

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