Question

$$\begin{gathered} \mathrm{Express}\mathrm{the}\mathrm{sum}3{\mathrm{e}}^4+3{\mathrm{e}}^9+3{\mathrm{e}}^{16}+4{\mathrm{e}}^4+4{\mathrm{e}}^9+4{\mathrm{e}}^{16} \\ \mathrm{using}\mathrm{double}-\mathrm{summation}\mathrm{notation}. \end{gathered}$$

Short answer

$3{\mathrm{e}}^4+3{\mathrm{e}}^9+3{\mathrm{e}}^{16}+4{\mathrm{e}}^4+4{\mathrm{e}}^9+4{\mathrm{e}}^{16}=\sum _{\mathrm{j}=3}^{\mathrm{j}=4}\sum _{\mathrm{k}=2}^{\mathrm{k}=4}{\mathrm{je}}^{{\mathrm{k}}^2}$

Step-by-step solution

Step 1. Given Information

$3{\mathrm{e}}^4+3{\mathrm{e}}^9+3{\mathrm{e}}^{16}+4{\mathrm{e}}^4+4{\mathrm{e}}^9+4{\mathrm{e}}^{16}\mathrm{in}\mathrm{double}-\mathrm{summation}\mathrm{notation}.$

Step 2. Finding common terms and converting to summation.

$$\begin{gathered} 3{\mathrm{e}}^4+3{\mathrm{e}}^9+3{\mathrm{e}}^{16}+4{\mathrm{e}}^4+4{\mathrm{e}}^9+4{\mathrm{e}}^{16}=(3{\mathrm{e}}^4+3{\mathrm{e}}^9+3{\mathrm{e}}^{16})+(4{\mathrm{e}}^4+4{\mathrm{e}}^9+4{\mathrm{e}}^{16}) \\ \mathrm{The}\mathrm{common}\mathrm{term}\mathrm{is}{\mathrm{e}}^4+{\mathrm{e}}^9+{\mathrm{e}}^{16}\mathrm{and}\mathrm{ranges}\mathrm{from}3\mathrm{to}4,\mathrm{which}\mathrm{can}\mathrm{be}\mathrm{converted}\mathrm{as} \\ 3{\mathrm{e}}^4+3{\mathrm{e}}^9+3{\mathrm{e}}^{16}+4{\mathrm{e}}^4+4{\mathrm{e}}^9+4{\mathrm{e}}^{16}=(3{\mathrm{e}}^4+3{\mathrm{e}}^9+3{\mathrm{e}}^{16})+(4{\mathrm{e}}^4+4{\mathrm{e}}^9+4{\mathrm{e}}^{16}) \\ =\sum _{\mathrm{j}=3}^4{\mathrm{je}}^4+{\mathrm{je}}^9+{\mathrm{je}}^{16} \\ =\sum _{\mathrm{j}=3}^4{\mathrm{je}}^{2^2}+{\mathrm{je}}^{3^2}+{\mathrm{je}}^{4^2} \\ \mathrm{The}\mathrm{common}\mathrm{term}\mathrm{is}{\mathrm{e}}^{2^2}\mathrm{and}\mathrm{ranges}\mathrm{from}2\mathrm{to}4,\mathrm{which}\mathrm{can}\mathrm{be}\mathrm{converted}\mathrm{as} \\ 3{\mathrm{e}}^4+3{\mathrm{e}}^9+3{\mathrm{e}}^{16}+4{\mathrm{e}}^4+4{\mathrm{e}}^9+4{\mathrm{e}}^{16}=\sum _{\mathrm{j}=3}^4{\mathrm{je}}^{2^2}+{\mathrm{je}}^{3^2}+{\mathrm{je}}^{4^2} \\ =\sum _{\mathrm{j}=3}^{\mathrm{j}=4}\sum _{\mathrm{k}=2}^{\mathrm{k}=4}{\mathrm{je}}^{{\mathrm{k}}^2} \\ \mathrm{Hence},\boxed{3{\mathrm{e}}^4+3{\mathrm{e}}^9+3{\mathrm{e}}^{16}+4{\mathrm{e}}^4+4{\mathrm{e}}^9+4{\mathrm{e}}^{16}=\sum _{\mathrm{j}=3}^{\mathrm{j}=4}\sum _{\mathrm{k}=2}^{\mathrm{k}=4}{\mathrm{je}}^{{\mathrm{k}}^2}} \end{gathered}$$