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Prove the identity: $$ \sum_{j=0}^n {n \choose j} 2^{n-j} {j \choose \left[\frac{j}{2}\right]} = {2n+1 \choose n} $$

In this particular problem, I tried to use induction and prove... But it is going too clumsy and lengthy... I need some sort of combinatorial arguments or something, to prove the claim in a elegant fashion.

Matti P.
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Maths lover
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  • Try it "combinatorially": interpret the right hand side as a choice of elements from a set (or better, the size of a certain set). Then, for each $j$, interpret a term in the left hand side as a choice or a size of a related set. Then make sure that every choice or counting in the left hand side sums to the total in the right hand side. – LetGBeTheGraph Dec 03 '20 at 13:18
  • @LetGBeTheGraph... Actually I am new to combinatorial arguments can you kindly provide atleast some answer in the answer section... As it is difficult for me to grasp like that – Maths lover Dec 03 '20 at 13:20
  • Another proof is also given here: https://math.stackexchange.com/questions/3115054/ – player3236 Dec 03 '20 at 13:25

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