Question

Evaluate the sums in Exercises $23–28$.

$\sum _{j=1}^m\sum _{k=1}^n(j-k)$

Short answer

The value of the summation is$\frac{1}{2}mn(m-n)$

Step-by-step solution

Step 1. Given information

An expression is given as$\sum _{j=1}^m\sum _{k=1}^n(j-k)$

Step 2. Evaluating summation

First separate out the term then simplify,

$$\begin{gathered} \sum _{j=1}^m\sum _{k=1}^m(j-k)=\sum _{j=1}^m\left(\sum _{k=1}^n(j-k)\right) \\ \sum _{k=1}^n(j-k)=(j-1)+(j-2)+(j-3)+\dots +(j-n) \\ =n\times j-(1+2+3+\dots +n) \\ =n\times j-\frac{n(n+1)}{2} \end{gathered}$$

Now simplify the other terms in the same manner,

$$\begin{gathered} \sum _{j=1}^m\sum _{k=1}^n(j-k)=\sum _{j=1}^m\left(\sum _{k=1}^n(j-k)\right) \\ =\sum _{j=1}^m\left(n\times j-\frac{n(n+1)}{2}\right) \\ =\left(n\times 1-\frac{n(n+1)}{2}\right)+\left(n\times 2-\frac{n(n+1)}{2}\right)+\left(n\times 3-\frac{n(n+1)}{2}\right)+.......+\left(n\times m-\frac{n(n+1)}{2}\right) \\ =n(1+2+3+\dots +m)-m\times \frac{n(n+1)}{2} \\ =n\times \frac{m(m+1)}{2}-m\times \frac{n(n+1)}{2} \\ =mn\left(\frac{m+1}{2}-\frac{n+1}{2}\right) \\ =mn\left(\frac{m-n}{2}\right) \\ =\frac{1}{2}mn(m-n) \end{gathered}$$