Question

Compute each of these double sums

$\begin{array}{r}\sum _{\mathrm{i}=1}^3 \sum _{\mathrm{j}=1}^2 (\mathrm{i}-\mathrm{j})\\ \sum _{\mathrm{i}=0}^3 \sum _{\mathrm{j}=0}^2 (3\mathrm{i}+2\mathrm{j})\\ \sum _{\mathrm{i}=1}^3 \sum _{\mathrm{j}=0}^2 \mathrm{j}\\ \sum _{\mathrm{i}=0}^2 \sum _{\mathrm{j}=0}^3 {\mathrm{i}}^2{\mathrm{j}}^3\end{array}$

Short answer

Thus, the double sums are given below:

1. 3
2. 78
3. 9
4. 180

Step-by-step solution

Step 1:

Given: $\sum _{i=1}^3 \sum _{j=1}^2 (i-j)$

Compute this double sum by first computing the inner sum followed by the outer sum.

$\begin{array}{r}\sum _{\mathrm{i}=1}^3 \sum _{\mathrm{j}=1}^2 (\mathrm{i}-\mathrm{j})\\ =\sum _{\mathrm{i}=1}^3 (\mathrm{i}-1+\mathrm{i}-2)\\ =\sum _{\mathrm{i}=1}^3 (2\mathrm{i}-3)\\ =2\left(1\right)-3+2\left(2\right)-3+2\left(3\right)-3\\ =3\end{array}$

Step 2:

Given:$\sum _{\mathrm{i}=0}^3 \sum _{\mathrm{j}=0}^2 (3\mathrm{i}+2\mathrm{j})$

Compute this double sum by first computing the inner sum followed by the outer sum.

role="math" $\begin{array}{r}\sum _{i=0}^3 \sum _{j=0}^2 (3i+2j)\\ =\sum _{i=0}^3 (3i+0+3i+2+3i+4)\\ =\sum _{i=0}^3 (9i+6)\\ =9\left(0\right)+6+9\left(1\right)+6+9\left(2\right)+6+9\left(3\right)+6\\ =78\end{array}$

Step 3:

Given:$\sum _{i=1}^3 \sum _{j=0}^2 j$

Compute this double sum by first computing the inner sum followed by the outer sum.

$\begin{array}{r}\sum _{i=1}^3 \sum _{j=0}^2 j\\ =\sum _{i=1}^3 (0+1+2)\\ =\sum _{i=1}^3 \left(3\right)\\ =3+3+3\\ =9\end{array}$

Step 4:

Given:$\sum _{i=0}^2 \sum _{j=0}^3 j^2{j}^3$

Compute this double sum by first computing the inner sum followed by the outer sum.

$\begin{array}{r}\sum _{\mathrm{i}=0}^2 \sum _{\mathrm{j}=0}^3 {\mathrm{i}}^2{\mathrm{j}}^3\\ =\sum _{\mathrm{i}=0}^2 \left(0+{\mathrm{i}}^2+8{\mathrm{i}}^2+27{\mathrm{i}}^2\right)\\ =\sum _{\mathrm{i}=0}^2 \left(36{\mathrm{i}}^2\right)\\ =36(0+1+4)\\ =180\end{array}$