Summation #6

Consider the Maclaurin series of $\sin x$ ,

$\begin{aligned} \sum_{n=0}^\infty \frac {(-1)^n x^{2n+1}}{(2n+1)!} & = \sin x & \small \color{#3D99F6} \text{Integrate both sides w.r.t. }x \\ \sum_{n=0}^\infty \frac {(-1)^n x^{2n+2}}{(2n+2)!} & = \color{#D61F06} - \cos x \color{#3D99F6} + 1 & \small \color{#3D99F6} \text{Since }\cos 0 = 1 \\ \sum_{n=0}^\infty \frac {(-1)^n x^{2n+2}}{(2n+2)!} & = {\color{#D61F06} - 1 + 2 \sin^2 \frac x2} + 1 & \small \color{#3D99F6} \text{Let }x = \sqrt{\frac 32} \\ \sum_{n=0}^\infty \frac {(-1)^n 3^{n+1}}{2^{n+1}(2n+2)!} & = 2 \sin^2 \left(\frac {\sqrt {\frac 32}}2\right) & \small \color{#3D99F6} \text{Multiply both sides by }\frac 1{18} \\ \sum_{n=0}^\infty \frac {(-1)^n 3^{n-1}}{2^{n+2}(2n+2)!} & = \frac 19 \sin^2 \left(\frac {\sqrt 6}4\right) \end{aligned}$

Therefore, $a+b+c = 6+4+9 = \boxed{19}$ .