1+x22!+x44!+…….∞2=

1+x22+x44+…….∞2 =ex+e-x22 =14e2x+e-2x2+1 =121+(2x)44+...+1 =1+2x22+23x44+...

1+x22!+x44!+…….∞2=

1+x22+x44+…….∞2 =ex+e-x22 =14e2x+e-2x2+1 =121+(2x)44+...+1 =1+2x22+23x44+...

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${(1 + \frac{x^{2}}{2!} + \frac{x^{4}}{4!} + \dots \dots . \infty)}^{2} =$

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$e^{2 x} + e^{- 2 x}$

$1 + \frac{2 x^{2}}{2!} + \frac{2^{3} x^{4}}{4!} + \frac{2^{3} x^{6}}{6!} + \dots \dots . \infty$

$1 + \frac{2 x^{2}}{2!} + \frac{2^{3} x^{4}}{4!} + \frac{2^{5} x^{6}}{6!} + \dots \dots . \infty$

$1 + \frac{x^{2}}{1!} + \frac{2 x^{2}}{2!} + \frac{3 x^{3}}{3!} + \dots \dots . \infty$

answer is A.

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Detailed Solution

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