Proving $\sum_\limits{k=0}^n \binom {k+m} {k} = \binom {m+n+1} {n}$ by induction.

Asked Nov 16 '16 at 20:19

Active Nov 16 '16 at 21:21

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$\sum_\limits{k=0}^n \binom {k+m} {k} = \binom {m+n+1} {n}$, where $n, m \in \Bbb N$.

I missed two lectures due to illness and now have this for homework. I am completely out of my depth, mostly due to there being a second variable to deal with.

edited Nov 16 '16 at 20:42

Jose M Serra

2,583

asked Nov 16 '16 at 20:19

Dystr

Well, first off, you can't start $k$ from $0$ if you then divide by it. Also, is $m$ a constant? – Riccardo Orlando Nov 16 '16 at 20:21
It's a copy paste of what's on the sheet, double checked it. I would assume m to be a constant. I am fairly new to the topic, but there is nothing determining any behavior for m. – Dystr Nov 16 '16 at 20:23
Does this mean I should simply state that the claim is false due to the division by 0? – Dystr Nov 16 '16 at 20:25
Well, that's rather perplexing. Also, the statement is false for $n=1$, and skipping the $k=0$ term. – Riccardo Orlando Nov 16 '16 at 20:25
Oh lord, they were binomials. Ok, now it makes sense! – Riccardo Orlando Nov 16 '16 at 20:26
@RiccardoOrlando: The OP has no idea what he is writing there: those are binomial coefficients (therefore without fraction line), not fractions. – Alex M. Nov 16 '16 at 20:26
Skipping those lectures really did hurt, didn't it? ;) – Riccardo Orlando Nov 16 '16 at 20:27
Well, this will be a fun semester.. I didn't chose to, I was simply too sick to go. – Dystr Nov 16 '16 at 20:27
Cheer up! Look up the definition for the binomial, and you'll see that the induction is actually doable! – Riccardo Orlando Nov 16 '16 at 20:29
Will do. One thing: Is there anything special about how to deal with the constant in this case? For some reason it really put me off. – Dystr Nov 16 '16 at 20:30
Nop, just treat it like any ol' number. That's the trick with most of the ugly looking things in maths: just don't look at them, until you've grown confident and they don't look so scary anymore.
In fact, you can do the very same proof and just write, say, 17 instead of m, and it'll probably work the same.
– Riccardo Orlando Nov 16 '16 at 20:31
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In that case, thank you very much. I might pester you again later, but for now I have a lead to work with! – Dystr Nov 16 '16 at 20:33
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You’ll find several proofs in answer to this question (and the one of which it was already a duplicate); this one is by induction. – Brian M. Scott Nov 16 '16 at 20:37
I don't claim to be certain of this, but I don't see how those overlap. Then again, I'm a total noob so I will trust your judgment. – Dystr Nov 16 '16 at 20:41
MathJax tip: use \sum_{k=0} instead of \sum\limits{k=0} – Simply Beautiful Art Nov 16 '16 at 21:24

Proving $\sum_\limits{k=0}^n \binom {k+m} {k} = \binom {m+n+1} {n}$ by induction.

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