e7x=1+7x+(7x)22+(7x)33+⋯⋯∞e-7x=1-7x+(7x)22-(7x)33+⋯⋯e7x-e-7x=27x+(7x)33+(7x)55+⋯⋯

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$e^{7 x} - e^{- 7 x} =$

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$1 + \frac{7 x}{1!} + \frac{{(7 x)}^{2}}{2!} + \dots \dots$

$1 + \frac{{(7 x)}^{2}}{2!} + \frac{{(7 x)}^{4}}{4!} + \dots \dots$

$2 (1 + \frac{{(7 x)}^{2}}{2!} + \frac{{(7 x)}^{4}}{4!} + \dots \dots)$

$2 (1 + \frac{7 x}{1!} + \frac{{(7 x)}^{3}}{3!} + \frac{{(7 x)}^{5}}{5!} + \dots \dots)$

answer is D.

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Detailed Solution

$\begin{array}{rcl} e^{7 x} & = & 1 + 7 x + \frac{{(7 x)}^{2}}{2} + \frac{{(7 x)}^{3}}{3} + \dots \dots \infty \\ e^{- 7 x} & = & 1 - 7 x + \frac{{(7 x)}^{2}}{2} - \frac{{(7 x)}^{3}}{3} + \dots \dots \\ e^{7 x} - e^{- 7 x} & = & 2 [7 x + \frac{{(7 x)}^{3}}{3} + \frac{{(7 x)}^{5}}{5} + \dots \dots] \end{array}$