A triple summation problem: $\sum_{i=1}^{100} \sum_{j=i}^{100} \sum_{k=i}^{j} 1$

Question

$$\begin{align}\sum_{i=1}^{100} \sum_{j=i}^{100} \sum_{k=i}^{j} 1&= \sum_{i=1}^{100} \sum_{j=i}^{100} (j-i+1)\\&= \sum_{i=1}^{100} \left(\sum_{j=i}^{100} j - i \sum_{j=i}^{100} 1 + \sum_{j=i}^{100} 1\right)\\&= \sum_{i=1}^{100} (5050 - 100i + 100)\\&= \sum_{i=1}^{100} (5150 - 100i)\\&= 5150 \sum_{i=1}^{100} 1 - 100 \sum_{i=1}^{100} i\\&= 515000 - 505000\\&= 10000\end{align}$$

I think the answer should be $171700$, but I can't figure out where I made a mistake

Note: Today I came across a computer science problem that requires this calculation, and, just a few hours ago, I had no idea how to do a double summation.

So yeah, your help would be needed.

In third equation you are expanding $\sum_{j=1}^{100}$ inside brackets instead of $\sum_{j=i}^{100}$ — Anton Grudkin, May 16 '24 at 14:17

score 1 · Answer 1 · answered May 16 '24 at 14:20

1

Hint $$\sum_{j=i}^{100} j=i+(i+1)+...100\\=\frac{100×101}{2} -\frac{(i-1)i}{2}$$

answered May 16 '24 at 14:20

Gwen

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A triple summation problem: $\sum_{i=1}^{100} \sum_{j=i}^{100} \sum_{k=i}^{j} 1$

1 Answers1