# It took me a while to solve this!

$\large \sum_{k=0}^{m} \,\,\large \frac{{2n-k}\choose{k}}{{2n-k}\choose{n}} \cdot \ \frac{2n-4k+1}{2n-2k+1} \cdot \,2^{n-2k}$

Given that $$n$$ is a positive integer and $$m$$ and $$n$$ are non-negative integers such that $$m \,\leq\, n$$, which of the given option is the closed form expression of given summation above.

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