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I would appreciate if somebody could help me with the following problem

Q: Find ? $(n,k\in\mathbb{N},k\leq n)$ $$\binom{n}{0} +\binom{n}{1}+\binom{n}{2}+\ldots+\binom{n}{k}=\,?$$

Michael Hardy
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Young
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    Refer the following. http://www.proofwiki.org/wiki/Sum_of_Binomial_Coefficients_for_Given_n –  Dec 07 '13 at 02:27
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    There is no closed form, see here: http://mathoverflow.net/questions/17202/sum-of-the-first-k-binomial-coefficients-for-fixed-n – universalset Dec 07 '13 at 02:28
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    @LinearAlgebra observe that $k$ and $n$ are different in the asker's question. – universalset Dec 07 '13 at 02:28
  • There is no closed form for partial sums . – abkds Dec 07 '13 at 02:29
  • @universalset: Thanks for pointing that out, I misunderstood the question. :) –  Dec 07 '13 at 02:30
  • The phrase "there is not closed form" should be followed by the predicate (gram.) "as a sum, with a fixed number of summands, of hypergeometric functions". If you allow more possible writings then there are closed forms. – OR. Dec 07 '13 at 02:34
  • Partial sum of rows of Pascal triangle doesn't have a closed-form expression. http://math.stackexchange.com/questions/69532/partial-sum-of-rows-of-pascals-triangle – Alex Dec 07 '13 at 02:29

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I think that I found a closed form which write
2^n - Binomial[n, 1 + k] Hypergeometric2F1[1, 1 + k - n, 2 + k, -1]
I hope I am not wrong.

Claude Leibovici
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