Question

Evaluate the sums in Exercises $23–28$.
$\sum _{j=1}^3\sum _{k=1}^4(3j-4k)$

Short answer

The value of the summation is$-48$

Step-by-step solution

Step 1. Given information

An expression is given as $\sum _{j=1}^3\sum _{k=1}^4(3j-4k)$

Step 2. Evaluating summation

First separate out the term and then simplify,

$$\begin{gathered} \sum _{j=1}^3\sum _{i=1}^4(3j-4k)=\sum _{j=1}^3\left(\sum _{k=1}^4(3j-4k)\right) \\ \sum _{k=1}^4(3j-4k)=(3j-4\times 1)+(3j-4\times 2)+(3j-4\times 3)+(3j-4\times n) \\ =12j-4\times (1+2+3+4) \\ =12j-4\times 10 \\ =12j-40 \\ \sum _{j=1}^3\sum _{k=1}^4(3j-4k)=\sum _{j=1}^3\left(\sum _{k=1}^4(3j-4k)\right) \\ =\sum _{j=1}^3(12j-40) \\ =(12\times 1-40)+(12\times 2-40)+(12\times 3-40) \\ =12(1+2+3)-3\times 40 \\ =12\times 6-120 \\ =72-120 \\ =-48 \end{gathered}$$