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Evaluate the following summation:

$$\large\sum_{i_1\le i_2\le i_3\le \cdots \le i_m}^n \left[\prod_{r=1}^m (i_r+2r-2)\right]\bigg /(2m-1)!!\\ =\large\sum_{i_1=1}^n\sum_{i_2=i_1}^n\sum_{i_3=i_2}^n\cdots\sum_{i_m=i_{m-1}}^n\frac {i_1(i_2+2)(i_3+4)(i_4+6)\cdots(i_m+2m-2)}{1\cdot 3\cdot 5\cdot 7\cdots (2m-1)}$$

Background
This is the generalized case of the problem posted here earlier. Please excuse the large font - this is for clearer display of subscripts of subscripts.

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$$\begin{align} &\sum_{i_1\le i_2\le i_3\le \cdots \le i_m}^n \left[\prod_{r=1}^m (i_r+2r-2)\right]\bigg /(2m-1)!!\\ &=\sum_{i_1=1}^n\sum_{i_2=i_1}^n\sum_{i_3=i_2}^n\cdots\sum_{i_m=i_{m-1}}^n\frac {i_1(i_2+2)(i_3+4)(i_4+6)\cdots(i_m+2m-2)}{1\cdot 3\cdot 5\cdot 7\cdots (2m-1)}\\ &=\sum_{i_m=1}^n\sum_{i_{m-1}=1}^{i_m}\cdots\sum_{i_3=1}^{i_4}\sum_{i_2=1}^{i_3}\sum_{i_1=1}^{i_2}\frac {i_1(i_2+2)(i_3+4)(i_4+6)\cdots(i_m+2m-2)}{1\cdot 3\cdot 5\cdot 7\cdots (2m-1)}\\ &=\sum_{i_m=1}^n\frac{i_m+2m-2}{2m-1}\sum_{i_{m-1}=1}^{i_m}\frac{i_{m-1}+2m-4}{2m-3}\cdots\sum_{i_3=1}^{i_4}\frac{i_3+4}5\sum_{i_2=1}^{i_3}\frac{i_2+2}3\sum_{i_1=1}^{i_2}\frac{i_1}1\\ &=\sum_{i_m=1}^n\frac{i_m+2m-2}{2m-1}\sum_{i_{m-1}=1}^{i_m}\frac{i_{m-1}+2m-4}{2m-3}\cdots\sum_{i_3=1}^{i_4}\frac{i_3+4}5\sum_{i_2=1}^{i_3}\frac{i_2+2}3\binom{i_2+1}2\\ &=\sum_{i_m=1}^n\frac{i_m+2m-2}{2m-1}\sum_{i_{m-1}=1}^{i_m}\frac{i_{m-1}+2m-4}{2m-3}\cdots\sum_{i_3=1}^{i_4}\frac{i_3+4}5\sum_{i_2=1}^{i_3}\binom{i_2+2}3\\ &=\sum_{i_m=1}^n\frac{i_m+2m-2}{2m-1}\sum_{i_{m-1}=1}^{i_m}\frac{i_{m-1}+2m-4}{2m-3}\cdots\sum_{i_3=1}^{i_4}\frac{i_3+4}5\binom{i_3+3}4\\ &=\sum_{i_m=1}^n\frac{i_m+2m-2}{2m-1}\sum_{i_{m-1}=1}^{i_m}\frac{i_{m-1}+2m-4}{2m-3}\cdots\sum_{i_3=1}^{i_4}\binom{i_3+4}5\\ &=\vdots\\ &=\sum_{i_m=1}^n\frac{i_m+2m-2}{2m-1}\sum_{i_{m-1}=1}^{i_m}\frac{i_{m-1}+2m-4}{2m-3}\binom{i_{m-1}+2m-5}{2m-4}\\ &=\sum_{i_m=1}^n\frac{i_m+2m-2}{2m-1}\sum_{i_{m-1}=1}^{i_m}\binom{i_{m-1}+2m-4}{2m-3}\\ &=\sum_{i_m=1}^n\frac{i_m+2m-2}{2m-1}\binom{i_{m}+2m-3}{2m-2}\\ &=\sum_{i_m=1}^n\binom{i_{m}+2m-2}{2m-1}\\ &=\binom{n+2m-1}{2m}\qquad\blacksquare\\ \end{align}$$


NB: for the case where $m=3$, as in the question here, this becomes $$\sum_{i_1=1}^n\sum_{i_2=i_1}^n\sum_{i_3=i_2}^n\frac {i_1(i_2+2)(i_3+4)}{1\cdot 3\cdot 5}=\binom {n+5}6$$

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