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

Question

How would I go about solving this summation?

$$\sum_{j=i+1}^n (j-i)$$

So far I have:

Using variable substitution

let ${k = j - i}$

$$ \sum_{j=i+1}^n (j-i) = \sum_{k=1}^{n-i} k$$

$$S = 1 + 2 + 3 + ... + (n - i)-2 + (n - i) - 1 + (n - i) $$

$$S = (n-i) + (n-i)-1 + (n-i)-2 + ... + 3 + 2 + 1$$

$$2S = (n-i+1)(n-i)$$

$$S = 1/2(((n-i+1)(n-i))) $$

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

Am I following this right?

Answer 1

• Set $k = j-i$, so replace all uses of $j$ with $k+i$.
• For the lower bound, subtract $i$ from both sides of the assignment.
• For the upper bound, subtract $i$. ("... to $k+i = n$ ...".)
• In the summand, simplify $k+i-i$.

This should leave a sum you know how to evaluate.

Answer 2

Following your approach by splitting the sum you can use $\sum_{k=1}^m k = \frac 12 m(m+1)$ twice as follows:

\begin{eqnarray*} \sum_{j=i+1}^n (j-i) & = & \sum_{j=i+1}^n j - \sum_{j=i+1}^n i \\ & = & \underbrace{\sum_{j=i+1}^n j + \sum_{j=1}^{\color{blue}{i}} j}_{\sum_{\color{blue}{j=1}}^n j}- \sum_{j=1}^{\color{blue}{i}} j - i\sum_{j=i+1}^n 1 \\ & = & \sum_{\color{blue}{j=1}}^n j - \sum_{j=1}^{\color{blue}{i}} j - i\sum_{j=i+1}^n 1 \\ & = & \frac 12 n(n+1) - \frac 12 i(i+1) - i(n-i)\\ & \stackrel{some \, algebra}{=} & \frac 12(n-i)(n+1-i) \end{eqnarray*}