Summation*summation = summation

$\begin{aligned} S & = \left[\left(\sum_{n=0}^\infty \frac {(-1)^nx^{2n}}{(2n+1)!4^n}\right)\left(\sum_{n=0}^\infty \frac {(-1)^nx^{2n+1}}{(2n)!4^n}\right)\right]^2 \\ & = \left[\left(\frac 2x \sum_{n=0}^\infty \frac {(-1)^n}{(2n+1)!}\left(\frac x2\right)^{2n+1}\right) \left( x \sum_{n=0}^\infty \frac {(-1)^n}{(2n)!}\left(\frac x2\right)^{2n}\right)\right]^2 \\ & = \left[2\sin \frac x2 \cos \frac x2 \right]^2 = \left[\sin x \right]^2 = \sin^2 x \\ & = 1 - \cos^2 x \\ & = 1 - \left[\sum_{n=0}^\infty \frac {(-1)^nx^{2n}}{\color{#3D99F6}(2n)!}\right]^2 & \small \color{#3D99F6} \text{Note that } 1\cdot 3 \cdot 5 \cdot ... \cdot (2n-1) = \frac {(2n)!}{2^nn!} \\ & = {\color{#D61F06}1} - \left[\sum_{n=0}^\infty \frac {(-1)^nx^{{\color{#D61F06}2}n}}{\color{#3D99F6} {\color{#D61F06}2}^n ({\color{#D61F06}1}n)! \prod_{k=1}^n (2k-1)}\right]^2 \end{aligned}$

$$

$\implies \dfrac {(a+b)(c+d)}3 = \dfrac {(1+2)(2+1)}3 = \boxed{3}$

$\displaystyle S = \left[\left(\sum_{n=0}^\infty \left(\frac{-1}{4}\right)^n \frac{x^{2n}}{(2n+1)!}\right)\left(\sum_{n=0}^\infty \left(\frac{-1}{4}\right)^n \frac{x^{2n+1}}{(2n)!}\right)\right]^2 \\ =\displaystyle \left[\left(\sum_{n=0}^\infty \left(\frac{-1}{\color{#3D99F6}2^2}\right)^n \frac{x^{2n}}{(2n+1)!}\right)\left(\sum_{n=0}^\infty \left(\frac{-1}{\color{#3D99F6}2^2}\right)^n \frac{x^{2n}\cdot {\color{#3D99F6}x}}{(2n)!}\right)\right]^2 \\ = \displaystyle \left[\left(\sum_{n=0}^\infty \frac{(-1)^n x^{2n}\cdot{\color{#3D99F6}x}}{2^{2n}(2n+1)!}\right)\left(\sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{2^{2n}(2n)!}\right)\right]^2 \\ = \displaystyle \left[\left(\sum_{n=0}^\infty \frac 22 \cdot \frac{(-1)^n x^{2n+1}}{2^{2n}(2n+1)!}\right)\left(\sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{2^{2n}(2n)!}\right)\right]^2 \\ = \displaystyle \left[2\left(\sum_{n=0}^\infty \frac{(-1)^n \left(\frac{x}{2}\right)^{2n+1}}{(2n+1)!}\right)\left(\sum_{n=0}^\infty \frac{(-1)^n \left(\frac{x}{2}\right)^{2n}}{(2n)!}\right)\right]^2 \\ \text{The above summations are the Taylor Series expansions for the sine and cosine function respectively. Therefore, } \\ = \left[\displaystyle 2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)\right]^2 \quad\quad \small \color{#3D99F6} \text{By the trigonometric identity, } 2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right) = \sin x \\ \implies S = \sin^2 x = \boxed{1 - \cos^2 x} \\ \text{Writing the Taylor Series expansion of cos, we get - } \\ S = \displaystyle 1 - \left(\sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{(2n)!}\right)^2 \\ = \displaystyle 1 - \left(\sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{\prod_{k=1}^n (2k)\prod_{k=1}^n (2k-1)}\right)^2 \quad\quad \small \color{#3D99F6} \text{see note 1} \\ = \displaystyle 1 - \left(\sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{{\color{#3D99F6}2^n n!}\prod_{k=1}^n (2k-1)}\right)^2 \quad\quad \small \color{#3D99F6} \text{see note 2} \\ = \displaystyle {\color{#D61F06}1} - \left(\sum_{n=0}^\infty \left(\frac{-1}{\color{#D61F06}2}\right)^n \frac{x^{{\color{#D61F06}2}n}}{({\color{#D61F06}1}n)!\prod_{k=1}^n (2k-1)}\right)^2 , \small \text{where, } {\color{#D61F06}a} = 1, \ {\color{#D61F06}b} = 2, \ {\color{#D61F06}c} = 2 \text{ and } {\color{#D61F06}d} = 1 \\ \implies \displaystyle \left(\frac{({\color{#D61F06}a} + {\color{#D61F06}b})({\color{#D61F06}c} + {\color{#D61F06}d})}{3}\right) = \displaystyle \left(\frac{({\color{#D61F06}1} + {\color{#D61F06}2})({\color{#D61F06}2} + {\color{#D61F06}1})}{3}\right) \\ = \displaystyle \frac{3\cdot3}{3} = \boxed{3} \\$

Note 1 :

$(2n)! \text{ can be written as a product of n even numbers and n odd numbers by the formula } - \\ (2n)! = \displaystyle \prod_{k=1}^n(2k)\prod_{k=1}^n(2k-1)$

Note 2:

$\displaystyle \prod_{k=1}^n(2k)$ is the product of n even numbers, which can be represented as -

$(2\cdot1)(2\cdot2)(2\cdot3)...(2\cdot n)$ . Taking out 2 common from all n terms, we get -

$(2\cdot1)(2\cdot2)(2\cdot3)...(2\cdot n) = \boxed{2^n n!}$