Closed form of many repeated summations of n

Question

I was looking into double summations, then I thought of repeated summations. As of now, I am having difficulty simplifying, for example

$$\sum_{r=1}^8...\sum_{z=1}^y\sum_{n=1}^z\sum_{i=1}^n i$$

Note that:

$$\sum_{n=1}^z\sum_{i=1}^n i =\sum_{i=1}^1i +\sum_{i=1}^2i+\sum_{i=1}^3i ...$$

Would it be useful if I posted the numerical answers generated? I was looking to solve it algebraically.

Nonetheless, I would appreciate any help in simplifying the term.

Thank you for your time.

Answer 1

Following a comment by Catalin Zara above,

$$\sum_{r=1}^8...\sum_{z=1}^y\sum_{n=1}^z(\sum_{i=1}^n i)$$

can be reduced to:

$$\sum_{r=1}^8...\sum_{z=1}^y\sum_{n=1}^z(\sum_{i=1}^n\sum_{j=1}^i 1)$$

Following this post,

The above can be simplified into:

$$\binom{8+k-1}{k}$$

Where k is the number of iterations.

In general, for k iterations:

$$\sum_{r=1}^v...\sum_{z=1}^y\sum_{n=1}^z(\sum_{i=1}^n i) = \binom{v+k-1}{k}$$

Thanks you all for helping :)

Answer 2

If you take the time of computing the sums, you will have $$\sum_{i=1}^n i=\frac{1}{2} n (n+1)$$ $$\sum_{n=1}^z \sum_{i=1}^n i=\frac{1}{2}\sum_{n=1}^zn (n+1)=\frac{1}{6} z (z+1) (z+2)$$ $$\sum_{z=1}^y \sum_{n=1}^z \sum_{i=1}^n i=\frac{1}{6}\sum_{z=1}^y z (z+1) (z+2)=\frac{1}{24} y (y+1) (y+2) (y+3)$$ $$\sum_{y=1}^x \sum_{z=1}^y \sum_{n=1}^z \sum_{i=1}^n i=\frac{1}{24}\sum_{y=1}^x y (y+1) (y+2) (y+3)=\frac{1}{120} x (x+1) (x+2) (x+3) (x+4)$$ where you crecognize some well known patterns.