I am tring to prove the following equality :
$\displaystyle \sum_{k=0}^n {k+l \choose l} = {n+l+1 \choose l+1} $
However, I did not manage to find a proof... Do you have any ideas ?
Thanks !
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Induction on $n$. – Did Apr 29 '15 at 14:35
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It works ! Thank you ! – Vincent Apr 29 '15 at 14:39
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You are welcome. You might want to post your solution as an answer and even, possibly, to accept it after a while. – Did Apr 29 '15 at 14:40
2 Answers
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Solved by induction on $n$ :
Basis : For $n=0$, we have ${l \choose l}={0+l+1 \choose l+1} = 1$
Inductive step : Let's assume that, for some nonnegative $n$ : $\displaystyle \sum_{k=0}^n {k+l \choose l} = {n+l+1 \choose l+1} $
Then : $\displaystyle \sum_{k=0}^{n+1} {k+l \choose l} = {n+l+1 \choose l+1} + {n+l+1 \choose l}= {n+l+2 \choose l+1}$
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Note that $$\binom {r+a}a=\binom {r+a+1} {a+1}- \binom {r+a}{a+1}$$ Hence $$\begin{align} \sum_{k=0}^n {k+l \choose l} &= \sum_{k=0}^n {k+l+1\choose l+1}-{k+l\choose l+1}\\ &={n+l+1 \choose l+1}\qquad\blacksquare \end{align}$$ via telescoping, as ${l\choose l+1}=0$.
Hypergeometricx
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