Question

Evaluate the sums in Exercises $23–28$

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

Short answer

The value of summation is

$\frac{lm{n}^2(l+1)(m+1)(2m+1)(n+1{)}^2}{48}$

Step-by-step solution

Step 1. Given information

An expression s given as$\sum _{i=1}^l\sum _{j=1}^m\sum _{k=1}^{m}i{j}^2{k}^3$

Step 2. Evaluating sum

Separate the first to last as,

$\sum _{i=1}^l\sum _{j=1}^m\sum _{k=1}^{k}i{j}^2{k}^3=\sum _{i=1}^l\sum _{j=1}^m\left(\sum _{i=1}^{n}i{j}^2{k}^3\right)$

Now simplify the summation as,

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

Now simplify the each terms as above,

$$\begin{gathered} \sum _{k=1}^1\sum _{j=1}^m\sum _{k=1}^{n}i{j}^2{k}^3=\sum _{i=1}^1\sum _{j=1}^m\left(\sum _{k=1}^{n}i{j}^2{k}^3\right) \\ =\sum _{j=1}^1\left(\sum _{j=1}^{m}i{j}^2\frac{n^2(n+1{)}^2}{4}\right) \\ \sum _{j=1}^m{i}^2\frac{n^2(n+1{)}^2}{4}=i\frac{n^2(n+1{)}^2}{4}\left(\sum _{j=1}^m{j}^2\right) \\ =i\frac{n^2(n+1{)}^2}{4}\left(1^2+2^2+3^2+\dots +m^2\right) \\ =i\times \frac{n^2(n+1{)}^2}{4}\times \frac{m(m+1)(2m+1)}{6} \\ \sum _{i=1}^1\sum _{j=1}^m\sum _{k=1}^{n}i{j}^2{k}^3=\sum _{i=1}^1\left(\sum _{j=1}^{m}i{j}^2\frac{n^2(n+1{)}^2}{4}\right) \\ =\sum _{i=1}^l\left(i\times \frac{n^2(n+1{)}^2}{4}\times \frac{m(m+1)(2m+1)}{6}\right) \\ =\frac{n^2(n+1{)}^2}{4}\times \frac{m(m+1)(2m+1)}{6}\times \left(\sum _{i=1}^1\left(i\right)\right) \\ =\frac{n^2(n+1{)}^2}{4}\times \frac{m(m+1)(2m+1)}{6}\times (1+2+3+\dots +l) \\ =\frac{n^2(n+1{)}^2}{4}\times \frac{m(m+1)(2m+1)}{6}\times \frac{l(l+1)}{2} \\ =\frac{lm{n}^2(l+1)(m+1)(2m+1)(n+1{)}^2}{48} \end{gathered}$$