A Weird Sum To infinity

Note that $n^4 + 3n^2 + 10n + 10 = (n^4 + 4) + (3n^2 + 10n + 6)$ , then $\large \frac {n^4 + 3n^2 + 10n + 10}{2^n (n^4+4)} = \frac1{2^n} + \frac{3n^2+10n+6}{2^n (n^4 + 4)}$ .

By Sophie Germain Identity, we can see that $n^4 + 4$ factors to $(n^2 - 2n+ 2)(n^2+2n+2)$ . Thus by partial fractions,

$\begin{aligned} \frac{3n^2+10n+6}{n^4 + 4} &\equiv & \frac4{n^2-2n+2} - \frac1{n^2+2n+2} \\ &=& \frac4{(n-1)^2+1} - \frac1{(n+1)^2 + 1} \end{aligned}$

Hence the series equals to

$\begin{aligned} && \displaystyle \sum_{n=2}^\infty \dfrac{n^4 + 3n^2 + 10n + 10}{2^n (n^4+4)} \\ &=& \displaystyle \sum_{n=2}^\infty \frac1{2^n} + \sum_{n=2}^\infty \left ( \frac1{2^n} \cdot \frac4{(n-1)^2+1} - \frac1{2^n} \cdot \frac1{(n+1)^2+1} \right ) \\ &=& \displaystyle \frac{1/2^2}{1-1/2} + \sum_{n=2}^\infty \left ( \frac1{2^{n-1}} \cdot \frac2{(n-1)^2+1} - \frac1{2^{n+1}} \cdot \frac2{(n+1)^2+1} \right ) \\ &=& \displaystyle \frac12 + \sum_{n=2}^\infty \left ( a_{n-1} - a_{n+1} \right) \\ &=& \displaystyle \frac12 + \left ( a_1 + a_2 \right) \\ &=& \frac12 + 2 \left( \frac1{2^1(1^2 + 1)} + \frac1{2^2(2^2+1)} \right) = \frac{11}{10} \\ \end{aligned}$

Hence, $a=11,b=10 \Rightarrow a(b+1) = \boxed{121}$ .