Crossover-101(BITSAT-2015)

This problem appeared in the BITSAT-2015 test that I wrote. I liked it, as it was one of the more interesting questions in the paper.

Using series expansion:

$A= \cos{ix}$ , where $i=\sqrt{-1}$

$\Rightarrow A^2=\cos^2{ix}=\frac{1+\cos{2ix}}{2}$

Use series expansion again on $\cos{2ix}$ to get option $3$ .

$We\quad know\quad that,\\ cosh(x)\quad =\quad \frac { { e }^{ x }+{ e }^{ -x } }{ 2 } \\ \quad \quad \quad \quad \quad \quad \quad =\quad 1+\frac { 1 }{ 2! } { x }^{ 2 }+\frac { 1 }{ 4! } { x }^{ 4 }+\frac { 1 }{ 6! } { x }^{ 6 }+...\\ \therefore cosh(1)\quad =\quad \sum _{ n=0 }^{ \infty }{ \frac { 1 }{ (2n)! } } =\quad { A }\quad \\ \therefore { (cosh(1)) }^{ 2 }\quad =\quad { A }^{ 2 }\quad =\quad { \left[ \frac { { e }^{ 1 }+{ e }^{ -1 } }{ 2 } \right] }^{ 2 }\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { { e }^{ 2 }+{ e }^{ -2 }+2 }{ 4 } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { 1 }{ 2 } +\frac { 1 }{ 4 } \left[ 1+2+\frac { { 2 }^{ 2 } }{ 2! } +\frac { { 2 }^{ 3 } }{ 3! } +\frac { { 2 }^{ 4 } }{ 4! } +... \right] +\frac { 1 }{ 4 } \left[ 1-2+\frac { { 2 }^{ 2 } }{ 2! } -\frac { { 2 }^{ 3 } }{ 3! } +\frac { { 2 }^{ 4 } }{ 4! } -... \right] \quad \quad \quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { 1 }{ 2 } +\frac { 1 }{ 2 } \left[ 1+\frac { { 2 }^{ 2 } }{ 2! } +\frac { { 2 }^{ 4 } }{ 4! } +... \right] \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad 1+\sum _{ n=1 }^{ \infty }{ \frac { { 2 }^{ 2n-1 } }{ (2n)! } } \quad \quad \quad \quad \quad$

I hope I am right. If wrong, do correct me!