Question

Evaluate the sums in Exercises $23–28$.

$\sum _{i=1}^4\sum _{j=1}^3\sum _{k=1}^{2}i{j}^2{k}^3$

Short answer

The value of summation is$1260$

Step-by-step solution

Step 1. Given information

An expression is given as$\sum _{i=1}^4\sum _{j=1}^3\sum _{k=1}^{2}i{j}^2{k}^3$

Step 2. Evaluating summation

Separate out the terms as and simplify,

$$\begin{gathered} \sum _{i=1}^4\sum _{j=1}^3\sum _{k=1}^{2}i{j}^2{k}^3=\sum _{i=1}^4\sum _{j=1}^3\left(\sum _{k=1}^{2}i{j}^2{k}^3\right) \\ \sum _{k=1}^{2}i{j}^2{k}^3=i{j}^2{1}^3+i{j}^2{2}^3 \\ =i{j}^2\left(1^3+2^3\right) \\ =i{j}^2(1+8) \\ =9i{j}^2 \end{gathered}$$

Now solve every term similarly,

$$\begin{gathered} \sum _{j=1}^4\sum _{j=1}^3\sum _{k=1}^{2}i{j}^2{k}^3=\sum _{j=1}^4\sum _{j=1}^3\left(\sum _{i=1}^2{j}^2{k}^3\right) \\ =\sum _{i=1}^4\left(\sum _{j=1}^{3}9i{j}^2\right) \\ \sum _{j=1}^{3}9i{j}^2=9i\left(\sum _{j=1}^3{j}^2\right) \\ =9i\left(1^2+2^2+3^2\right) \\ =9i(1+4+9) \\ =9i\times 14 \\ =126i \\ \sum _{i=1}^4\sum _{j=1}^3\sum _{k=1}^{2}i{j}^2{k}^3=\sum _{i=1}^4\left(\sum _{j=1}^{3}9i{j}^2\right) \\ =\sum _{i=1}^4\left(126i\right) \\ =126\times \left(\sum _{i=1}^4\left(i\right)\right) \\ =126\times (1+2+3+4) \\ =126\times 10 \\ =1260 \end{gathered}$$