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## Summations and Combinations

Prove the following :

\(\sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k {n-k \choose k} \cdot 2^{n-2k} = n + 1\)

\(\sum_{k=0}^n {2k \choose k}{2n-2k \choose n-k} = 4^n\)

\(\sum_{k=0}^{n} 2^k {n \choose k}{n-k \choose \lfloor(n-k)/2\rfloor} = {2n+1 \choose n}\)