e5x=1+(5x)+(5x)22+(5x)33+⋯⋯e-5x=1-5x+(5x)22-(5x)33+⋯⋯e5x+e-5x=21+(5x)22+(5x)44+⋯⋯

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$e^{5 x} + e^{- 5 x} =$

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

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

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

$2 (1 + \frac{{(5 x)}^{2}}{2!} + \frac{{(5 x)}^{4}}{4!} + \dots \dots)$

answer is D.

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Detailed Solution

$\begin{array}{rcl} e^{5 x} & = & 1 + (5 x) + \frac{{(5 x)}^{2}}{2} + \frac{{(5 x)}^{3}}{3} + \dots \dots \\ e^{- 5 x} & = & 1 - 5 x + \frac{{(5 x)}^{2}}{2} - \frac{{(5 x)}^{3}}{3} + \dots \dots \\ e^{5 x} + e^{- 5 x} & = & 2 [1 + \frac{{(5 x)}^{2}}{2} + \frac{{(5 x)}^{4}}{4} + \dots \dots] \end{array}$