-1

Hello everyone I am a high school student and I need help with the proof that $$C(n , 0) + C(n+1 , 1) + C(n+2 , 2) + ... + C(n + k , k) =C(n+k+1 , k).$$

Mars Plastic
  • 4,239
tantan
  • 3
  • What have you tried? – user113102 Mar 13 '20 at 15:55
  • I tried to use pascal identity and induction – tantan Mar 13 '20 at 15:59
  • No the questions aren't the same – tantan Mar 13 '20 at 16:08
  • @tantan If you take a look at how the two questions fit into Pascal's triangle, you will see that they are actually the same, because the triangle is symmetric. It also gives you a keyword to search for more answers: Hockey sick identity. – Arthur Mar 13 '20 at 16:11
  • To see that these are the same summations, denote your general term by $\binom{n+j}j$, note that this is equal to $\binom{n+j}n$ and your sum is equal to $\binom{n+k+1}{n+1}$, and substitute $k\mapsto n$, $t\mapsto n+j$, $n\mapsto n+k$ in the other question. – joriki Mar 13 '20 at 16:15
  • Can you write the solution for my question? – tantan Mar 13 '20 at 16:19
  • 2
    There are $17$ answers there (I just added the $17$th earlier), and I showed you how to apply them to your question. If you don't know how to perform substitutions, I'd recommend asking a question about that; you won't get far without being able to do substitutions, so you might as well learn it now. By the way, I only read your message by chance because I still had the page open. If you want people to get notified, you need to ping them, using @username. – joriki Mar 13 '20 at 17:06

1 Answers1

1

$$\sum_{m=0}^{k}\binom{n+m}{m}=\sum_{m=0}^{k}\binom{n+m}{n}$$ Setting $n+m \mapsto m$ and using Hockey-stick identity follows:

$$=\sum_{m=n}^{k+n}\binom{m}{n}=\binom{k+n+1}{n+1}=\binom{k+n+1}{k}$$

Which is the claim.