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Sadly $\sum ^n _{k=0} \binom{a}{k}$ does not have a simple formula, but I thought you could maybe use $\sum ^n _{k=0} \binom{n}{k} = 2^n$ and the partial integration equivalent to find sums like $\sum ^n _{k=0} k\binom{n}{k}$ using finite calculus.

In know that $\Delta \binom{n}{k} = \binom{n}{k-1}$, but this only works when k is constant.

This post might be related, although I was not able to understand how $\Delta^n \binom{n}{k}$ and $\sum _k (-1)^{n-k}\binom{n}{k}\binom{n+k}{n}$ are related.

Contest math
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    What do you mean by "a simple formula"? Formulas really don't get much simpler than $2^n$... – lulu Feb 19 '22 at 16:49
  • @lulu Oh I messed that one up. I edited it to what I wanted to say. – Contest math Feb 19 '22 at 17:06
  • Well, again, $\sum_{k=0}^nk\binom nk$ has a simple expression, obtained by differentiating the binomial expansion of $(1+x)^n$. – lulu Feb 19 '22 at 17:43
  • I am aware of that, but I am searching for a way to solve this using finite calculus, because a method to do so could be useful in solving similar problems. – Contest math Feb 19 '22 at 19:13

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