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I am trying to prove it by algebra.Reference

What i tried so far is to simplify the binomials and cancelling similar variables. I stuck at some point without any clue on this step :

\begin{align*} \sum_{j=k}^n \binom{j}{k} & = \sum_{j=k}^{n} \frac{j!}{k!(j-k)!} & \text{[using the formula for binomial(j, k)]}\\ & = \sum_{j=k}^{n} \frac{j(j-1) \cdot \ldots \cdot (k+1)k!}{k!(j-k)!} & \text{[writing out the factorial]} \end{align*}

I would like for an help to understand if there is any simple way to prove this identity.

N. F. Taussig
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Ben Cohen
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  • "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Christian E. Ramirez Apr 06 '23 at 10:24
  • The lower index SHOULD BE ZERO for the identity to hold. That is "j=0" not "j=k". – NoChance Apr 06 '23 at 13:32
  • that's equivalent to j=k as you can see here https://math.stackexchange.com/questions/1490794/proof-of-the-hockey-stick-zhu-shijie-identity-sum-limits-t-0n-binom-tk – Ben Cohen Apr 06 '23 at 16:12

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