Question

If the value of $\sum _{r=0}^{100}(r^2+4r+4)(r+1)!$ is $\left(a\right)!-b$, where $0\le b<10$, then the sum of digits of $a+b$, is

Answer 1

$\sum _{r=0}^{100}(r^2+4r+4)(r+1)!=\sum _{r=0}^{100}(r+2)(r+2)!=\sum _{r=0}^{100}(r+3-1)(r+2)!=\sum _{r=0}^{100}(r+3)!-\sum _{r=0}^{100}(r+2)!=(3!+4!+\cdots +103!)-(2!+3!+\cdots +102!)=\left(103\right)!-2\therefore a+b=105$
Hence sum of digits of $(a+b)$ is $6$