Arithmetic Series on Factorials

Calculus Level 3

an=1+6(n1)bn=1+21(n1)cn=202+102(n1)\begin{aligned} a_n&=1+6(n-1) \\ b_n&=1+21(n-1) \\ c_n&=202+102(n-1) \end{aligned}

Given the above, what is the value of

12!+13!+a14!+b15!+c16!+a27!+b28!+c29!+a310!+b311!+c312!+?\displaystyle \frac 1{2!}+\frac 1{3!}+\frac {a_1}{4!}+\frac {b_1}{5!}+\frac {c_1}{6!}+\frac {a_2}{7!}+\frac {b_2}{8!}+\frac {c_2}{9!}+\frac {a_3}{10!}+\frac {b_3}{11!}+\frac {c_3}{12!}+\cdots\, ?

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