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I have a test coming up tomorrow, and on one of the practice papers, this question showed up:

Show that for all $k \in \mathbb{N}$, and all $n \in \mathbb{N}\cup\{0\}$, $$\sum_{j=k}^{n+k}{j\choose k} = {n+k+1\choose k+1}$$

I don't know how to prove this at all. I know how to prove thing using mathematical induction, but it gets harder I find when the binomial theorem is involved, let alone multiple variables. Does anyone know how I'd go about answering this? The solutions weren't posted (and it's not a part of any assignment, so I'm just doing this for practice).

Roland Killian
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Elmer Cado
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  • You should not post links to images. I have typed out the question for you. – Roland Killian Oct 07 '21 at 20:15
  • Please insert the question using mathjax to format the formula. Here is an introduction: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference For instance, to get that binomial coefficient type $\binom j k$ or $\binom{j}{k}$ when the two entries are "longer tokens". This is shown as $$\binom jk\ .$$Would you please try to edit the question and insert the formula to be shown?! Do you know any formula relating $$\binom j{k+1}$$ with binomial coefficients having $k$ as the lower entry?! – dan_fulea Oct 07 '21 at 20:18
  • @dan_fulea Could you approve my edit? – Roland Killian Oct 07 '21 at 20:19

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