IIT-JEE 2010 · Paper 1, Mathematics question 54

Sums. $S_1=0$ (first term 0), and for $k\ge2$
$$S_k=\frac{(k-1)/k!}{1-1/k}=\frac1{(k-1)!}.$$

Terms. $k^2-3k+1=(k-1)(k-2)-1$, so for $k\ge3$
$$(k^2-3k+1)S_k=\frac1{(k-3)!}-\frac1{(k-1)!}>0,$$
and for $k=2$ the term is $-1$, with absolute value 1.

Telescoping.
$$\sum_{k=3}^{100}\left(\frac1{(k-3)!}-\frac1{(k-1)!}\right)=\frac1{0!}+\frac1{1!}-\frac1{98!}-\frac1{99!}=2-\frac{100}{99!}=2-\frac{100^2}{100!}.$$

Total. $\frac{100^2}{100!}+1+2-\frac{100^2}{100!}=3$.

Answer: 3