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If $\frac{e^{x}+e^{5 x}}{e^{3 x}}=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+\ldots$ then
the value of $2 a_{1}+2^{3} a_{3}+2^{5} a_{5}+\ldots$ is
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the value of $2 a_{1}+2^{3} a_{3}+2^{5} a_{5}+\ldots$ is
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Let $\frac{\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{5 \mathrm{x}}}{\mathrm{e}^{3 \mathrm{x}}}=\mathrm{a}_{0}+\mathrm{a}_{1} \mathrm{x}+\mathrm{a}_{2} \mathrm{x}^{2}+\mathrm{a}_{3} \mathrm{a}^{3}+\ldots .$
$\begin{aligned} &=\frac{\mathrm{e}^{\mathrm{x}}}{\mathrm{e}^{3 \mathrm{x}}}+\frac{\mathrm{e}^{5 \mathrm{x}}}{\mathrm{e}^{3 \mathrm{x}}}=\mathrm{a}_{0}+\mathrm{a}_{1} \mathrm{x}+\mathrm{a}_{2} \mathrm{x}^{2}+\ldots . \\=& \mathrm{e}^{-2 \mathrm{x}}+\mathrm{e}^{2 \mathrm{x}}=\mathrm{a}_{0}+\mathrm{a}_{1} \mathrm{x}+\mathrm{a}_{2} \mathrm{x}^{2}+\mathrm{a}_{3} \mathrm{x}^{3}+\ldots . \end{aligned}$
By using $\mathrm{e}^{\mathrm{x}}=1+\mathrm{x}+\frac{\mathrm{x}^{2}}{2 !}+\frac{\mathrm{x}^{3}}{3 !}+{ }_{--}--$ and
$\mathrm{e}^{-\mathrm{x}}=1-\mathrm{x}+\frac{\mathrm{x}^{2}}{2 !}-\frac{\mathrm{x}^{3}}{3 !}+---$
$\mathrm{e}^{-2 \mathrm{x}}+\mathrm{e}^{2 \mathrm{x}}=2\left[1+\frac{(2 \mathrm{x})^{2}}{21}+\frac{(2 \mathrm{x})^{4}}{41}+\ldots .\right]$
$$
\begin{array}{l}
=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} a^{3}+\ldots . \\
=a_{1}=a_{3}=a_{5}=\ldots=0
\end{array}
$$
Hence, $2 \mathrm{a}_{1}+2^{3} \mathrm{a}_{3}+2^{5} \mathrm{a}_{5}+\ldots .=0$
$\begin{aligned} &=\frac{\mathrm{e}^{\mathrm{x}}}{\mathrm{e}^{3 \mathrm{x}}}+\frac{\mathrm{e}^{5 \mathrm{x}}}{\mathrm{e}^{3 \mathrm{x}}}=\mathrm{a}_{0}+\mathrm{a}_{1} \mathrm{x}+\mathrm{a}_{2} \mathrm{x}^{2}+\ldots . \\=& \mathrm{e}^{-2 \mathrm{x}}+\mathrm{e}^{2 \mathrm{x}}=\mathrm{a}_{0}+\mathrm{a}_{1} \mathrm{x}+\mathrm{a}_{2} \mathrm{x}^{2}+\mathrm{a}_{3} \mathrm{x}^{3}+\ldots . \end{aligned}$
By using $\mathrm{e}^{\mathrm{x}}=1+\mathrm{x}+\frac{\mathrm{x}^{2}}{2 !}+\frac{\mathrm{x}^{3}}{3 !}+{ }_{--}--$ and
$\mathrm{e}^{-\mathrm{x}}=1-\mathrm{x}+\frac{\mathrm{x}^{2}}{2 !}-\frac{\mathrm{x}^{3}}{3 !}+---$
$\mathrm{e}^{-2 \mathrm{x}}+\mathrm{e}^{2 \mathrm{x}}=2\left[1+\frac{(2 \mathrm{x})^{2}}{21}+\frac{(2 \mathrm{x})^{4}}{41}+\ldots .\right]$
$$
\begin{array}{l}
=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} a^{3}+\ldots . \\
=a_{1}=a_{3}=a_{5}=\ldots=0
\end{array}
$$
Hence, $2 \mathrm{a}_{1}+2^{3} \mathrm{a}_{3}+2^{5} \mathrm{a}_{5}+\ldots .=0$
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