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So I was assigned this homework problem: $$\ {s \choose s} + {s+1 \choose s} +...+ {n \choose s} = {n+1 \choose s+1}$$ for all s and all $n \geq s$ I've tried to email both my professor and my TA and their explanations seem contradictory. My professor responded saying the statement I need to prove is "The formula is correct for $0 \leq s \leq n$." Whereas my TA told me I need to use induction on both variables and I'm not sure how to do that. Any help is appreciated!

JMoravitz
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EpicViiper
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  • It should indeed be proven for all values of $s$ and $n$ such that $0\leq s\leq n$. Since you are tasked with showing it for all values of $s$ and $n$ satisfying this, using double-induction is standard. – JMoravitz Apr 12 '19 at 23:19
  • Thank you for linking that proof! I'll try to work my way through that proof now and see if I can apply it to mine. – EpicViiper Apr 12 '19 at 23:32

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Use induction on $n$. Fix $s \leq n$. If the equation holds then it holds with $n$ cha nged to $n+1$ because of the following identity: $\binom {n+1} {s+1} +\binom {n+1} s=\binom {n+2} {s+1}$. You can verify this identity by writing the binomial coefficients in terms of factorials. To complete the proof you only have to verify the equation with $n$ changed to $n+1$ and $s=n+1$. But this just reduces to $1=1$!

Kavi Rama Murthy
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  • So when I do my base case, do I set n and s to be equal to 1? Or just s? – EpicViiper Apr 12 '19 at 23:35
  • When $n=1$ you have to consider the cases $s=0$ as well as $s=1$. The first leads to $1+1=2$ and the second one leads to $1=1$. – Kavi Rama Murthy Apr 12 '19 at 23:37
  • Oh okay that makes sense. So then in my inductive step, I would assume that the formula is true for some s such that $s \leq n$. I want to show that it works for any $s \leq n+1$ then so would I just do the algebra in the equation? I.e. the last term on the LHS is ${n+1 \choose s}$ and the right hand side is ${n+2 \choose s+1}$ – EpicViiper Apr 12 '19 at 23:44
  • Yes. You have to separate the cases $s \leq n$ and $s=n+1$ as I have done. – Kavi Rama Murthy Apr 12 '19 at 23:46
  • Thank you so much for the help! – EpicViiper Apr 12 '19 at 23:52