How to evaluate the sum $\sum_{j = i+1}^{n-1} j $?

Question

How would I go about solving this summation?

$$\sum\limits_{j = i+1}^{n-1} j $$

I'm trying to figure out how to solve this summation using the fact that $\sum\limits_{i = 1}^{k} i= \frac{k(k+1)}{2}$.

The answer I get with my Ti-89 is $-\frac{i^2}{2}-\frac{-i}{2}+\frac{n^2}{2}-\frac{n}{2}$

You have $i$ and $n$ in the sum, but $k$ in the answer. Please edit. To get $k(k+1)/2$, the sum should be from $1$ to $k$. — , Jan 25 '15 at 02:22
Also, there's a chance your question is a duplicate of Proof for formula for sum of sequence 1+2+3+…+n — , Jan 25 '15 at 02:25
Well it's for my algorithms class and i'm trying to solve a nested summation and we havn't done anything like this in class where we have the lower limit as such — user210375, Jan 25 '15 at 02:30

score 4 · Answer 1 · answered Jan 25 '15 at 02:26

4

$$1+2+...+i+(i+1)+...+(n-1)=\frac{(n-1)n}2\\ \frac{i(i+1)}2+(i+1)+...+(n-1)=\frac{(n-1)n}2\\ \sum_{j=i+1}^{n-1}j=\frac12((n-1)n-i(i+1))$$

answered Jan 25 '15 at 02:26

Empy2

score 1 · Answer 2 · answered Jan 25 '15 at 02:46

1

Assuming that $n$ is greater than $i$, we have $$\sum_{j=i+1}^{n-1}j=\sum_{j=1}^{n-1} j-\sum_{j=1}^i j=\binom{n}{2}-\binom{i+1}{2}.$$

answered Jan 25 '15 at 02:46

Paolo Leonetti

2 Answers2