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I cannot understand anything about this question. I get what Pascal's triangle is and I have seen the formula for adding combinations, and that's the only help I have seen but I do not know how to apply it to this problem. I understand how induction works, but I can't even understand the base case for this problem, even with hints. This is my first time dealing with combinations in this way (but I do understand principles behind combinatorics...) so any help, I will appreciate very much.

Fix a positive integer $r$. Use mathematical induction to show that $\binom {r} {r} + \binom {r+1} {r} + \binom {r+2} {r} +...+ \binom {n} {r}=\binom {n+1} {r+1}$.

I know that the base case should be $n=r$ but I cannot understand how the base case works. Also, I have to use Pascal's rule for the proof and not binomal coefficients formula. Please help?

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    The base case is "$\binom rr = \binom{r+1}{r+1}$". Does that seem like a true statement to you? And why is that? – 2017-01-23
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    It does, because both are equal to one. I was having trouble because I thought the base case should be equal to $\binom{r}{r} + \binom{r}{r}$, since $\binom{r}{r}$ is a constant and should be added to $\binom{n}{r}$ where $n=r$. – 2017-01-23
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    I can see how that would be confusing, but that's not how that summation notation is supposed to be interpreted. It means "Keep adding $1$ to the top index, stop when you reach $n$, and then add them all together". In the base case, we reach $n$ before even adding the first $1$, so there is only one term on the left side. – 2017-01-23
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    Oh, ok. I will have to remember that for this particular notation. I was just doing inductive proofs for sequences, so I naturally followed the same principle for those, like adding a constant. I see what you mean now. Thanks for clarifying that. – 2017-01-23

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  1. $${r \choose r}=1={r+1 \choose r+1}$$
  2. $${r \choose r}+...+{k\choose r}={k+1 \choose r+1}$$
  3. $${r \choose r}+...+{k\choose r}+{k+1\choose r}=\\ {k+1 \choose r+1}+{k+1\choose r}=\\ \text{[using Pascal's Rule]:}\\ ={k+2 \choose {r+1}} $$
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    Ok, that was much simpler than I was trying to make it. That, and Arthur's explanation, helped a lot. Thank you very much. – 2017-01-23