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What would be a simplified formula for $\displaystyle \sum_{m=1}^{N}\binom{m+k-1}{m}$ for a given number $k$ and any number $N$?

Shivam Tomar
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  • This is quite an interesting problem but you should show what you've done so far or else your post is prone to getting downvotes. – Prasun Biswas Apr 04 '15 at 04:57
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    @PrasunBiswas Exactly what do you expect a person who doesn't know how to solve this problem to show? – DanielV Apr 04 '15 at 05:13
  • If the problem was encountered in a class, what topics are you currently studying? If the problem was encountered in a book being read independently, in what section was the problem encountered? Have you tried anything? If so, what, and where did you get stuck? Context such as this helps someone give an appropriate answer. – robjohn Apr 04 '15 at 08:01
  • This summation actually occurs in this recent answer. Any relation to that question. – Marc van Leeuwen Apr 04 '15 at 09:32
  • $$\begin{align} \sum_{m=1}^{N}{m+k-1\choose m} &=\sum_{m=1}^{N}{m+k-1\choose k-1}\ &=\sum_{m=1}^{N}\left[{m+k\choose k}-{m+k-1\choose k}\right]\ &=\sum_{m=2}^{N+1}{m+k-1\choose k}-\sum_{m=1}^N{m+k-1\choose k}\ &={N+k\choose k}-{k\choose k}\ &={N+k\choose k}-1\ \end{align}$$ – Hypergeometricx Apr 04 '15 at 15:26
  • This should in fact have been closed as a duplicate, e.g., of this. – Brian M. Scott Apr 04 '15 at 15:32

2 Answers2

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Use the following formula: $$\displaystyle\binom{m+k-1}{m}=\binom{m+k}{m}-\binom{m+k-1}{m-1}$$

Proof:

\begin{align} \binom{m+k}{m}-\binom{m+k-1}{m-1}&=\dfrac{(m+k)!}{m!\:k!}-\dfrac{(m+k-1)!}{(m-1)!\:k!} \\ &=\dfrac{(m+k-1)!}{(m-1)!\:k!}\left(\dfrac{m+k}{m}-1\right) \\ &=\dfrac{(m+k-1)!}{m!\:(k-1)!} \\ &=\binom{m+k-1}{m} \end{align}

So \begin{align} \sum \limits_{m=1}^{N}\binom{m+k-1}{m}&=\sum \limits_{m=1}^{N}(\binom{m+k}{m}-\binom{m+k-1}{m-1}) \\ &=\sum \limits_{m=1}^{N}\binom{m+k}{m}-\sum \limits_{m=1}^{N}\binom{m+k-1}{m-1} \\ &=\sum \limits_{m=1}^{N}\binom{m+k}{m}-\sum \limits_{m=0}^{N-1}\binom{m+k}{m} \\ &=\sum \limits_{m=1}^{N}\binom{m+k}{m}-\sum \limits_{m=1}^{N-1}\binom{m+k}{m}-1 \\ &=\binom{N+k}{N}-1 \end{align}

Eugene Zhang
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Hint: Try using a Telescoping Series and the recursion for Pascal's Triangle $$ \begin{align} \sum_{m=1}^N\binom{m+k-1}{m} &=\sum_{m=1}^N\binom{m+k-1}{k-1}\\ &=\sum_{m=1}^N\left[\binom{m+k}{k}-\binom{m+k-1}{k}\right]\\ &=\binom{N+k}{k}-\binom{k}{k} \end{align} $$

robjohn
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