Mathematics - Binomial Theorem Question with Solution | TestHub

Let $\frac{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{5\mathrm{x}}}{{\mathrm{e}}^{3\mathrm{x}}}={\mathrm{a}}_0+{\mathrm{a}}_1\mathrm{x}+{\mathrm{a}}_2{\mathrm{x}}^2+{\mathrm{a}}_3{\mathrm{a}}^3+\dots .$ $\begin{array}{rl}& =\frac{{\mathrm{e}}^{\mathrm{x}}}{{\mathrm{e}}^{3\mathrm{x}}}+\frac{{\mathrm{e}}^{5\mathrm{x}}}{{\mathrm{e}}^{3\mathrm{x}}}={\mathrm{a}}_0+{\mathrm{a}}_1\mathrm{x}+{\mathrm{a}}_2{\mathrm{x}}^2+\dots .\\ =& {\mathrm{e}}^{-2\mathrm{x}}+{\mathrm{e}}^{2\mathrm{x}}={\mathrm{a}}_0+{\mathrm{a}}_1\mathrm{x}+{\mathrm{a}}_2{\mathrm{x}}^2+{\mathrm{a}}_3{\mathrm{x}}^3+\dots .\end{array}$ By using ${\mathrm{e}}^{\mathrm{x}}=1+\mathrm{x}+\frac{{\mathrm{x}}^2}{2!}+\frac{{\mathrm{x}}^3}{3!}+{}_{--}--$ and ${\mathrm{e}}^{-\mathrm{x}}=1-\mathrm{x}+\frac{{\mathrm{x}}^2}{2!}-\frac{{\mathrm{x}}^3}{3!}+---$ ${\mathrm{e}}^{-2\mathrm{x}}+{\mathrm{e}}^{2\mathrm{x}}=2\left[1+\frac{(2\mathrm{x}{)}^2}{21}+\frac{(2\mathrm{x}{)}^4}{41}+\dots .\right]$ $$\begin{array}{l}=a_0+a_{1}x+a_2{x}^2+a_3{a}^3+\dots .\\ =a_1=a_3=a_5=\dots =0\end{array}$$ Hence, $2{\mathrm{a}}_1+2^3{\mathrm{a}}_3+2^5{\mathrm{a}}_5+\dots .=0$