$$ \begin{aligned} & S_1=3+7+11+15+19+\ldots . . \\\\ & S_2=1+6+11+16+21+\ldots . . \end{aligned} $$

is :

Let $$\sum_\limits{n=0}^{\infty} \frac{\mathrm{n}^{3}((2 \mathrm{n}) !)+(2 \mathrm{n}-1)(\mathrm{n} !)}{(\mathrm{n} !)((2 \mathrm{n}) !)}=\mathrm{ae}+\frac{\mathrm{b}}{\mathrm{e}}+\mathrm{c}$$, where $$\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathbb{Z}$$ and $$e=\sum_\limits{\mathrm{n}=0}^{\infty} \frac{1}{\mathrm{n} !}$$ Then $$\mathrm{a}^{2}-\mathrm{b}+\mathrm{c}$$ is equal to ____________.

Let $$a_1=b_1=1$$ and $${a_n} = {a_{n - 1}} + (n - 1),{b_n} = {b_{n - 1}} + {a_{n - 1}},\forall n \ge 2$$. If $$S = \sum\limits_{n = 1}^{10} {{{{b_n}} \over {{2^n}}}} $$ and $$T = \sum\limits_{n = 1}^8 {{n \over {{2^{n - 1}}}}} $$, then $${2^7}(2S - T)$$ is equal to ____________.

Let $$\{ {a_k}\} $$ and $$\{ {b_k}\} ,k \in N$$, be two G.P.s with common ratios $${r_1}$$ and $${r_2}$$ respectively such that $${a_1} = {b_1} = 4$$ and $${r_1} < {r_2}$$. Let $${c_k} = {a_k} + {b_k},k \in N$$. If $${c_2} = 5$$ and $${c_3} = {{13} \over 4}$$ then $$\sum\limits_{k = 1}^\infty {{c_k} - (12{a_6} + 8{b_4})} $$ is equal to __________.