QJEEAdv10Ma2606: JEE Advanced 2010 Mathematics - Class 11 JEEAdv10Ma2606 Solution: Let = 1, 2,....., 100, denote the sum of the infinite geometric …

$\begin{aligned} & \text { Using } S_{\infty}=\frac{a}{1-r} \text {, we get } \\\\ & \qquad S_k=\left\{\begin{array}{cc}0, & k=1 \\\\ \frac{1}{(k-1)!}, & k \geq 2\end{array}\right.\end{aligned}$

$\begin{aligned} & \text { Now } \sum_{k=1}^{100}\left|\left(k^2-3 k+1\right) S_k\right|=\sum_{k=2}^{100}\left|\left(k^2-3 k+1\right)\right| \frac{1}{(k-1)!} \\\\ &=|-1|+\sum_{k=3}^{100} \frac{\left(k^2-1\right)+1-3(k-1)-2}{(k-1)!} \\\\ & \quad \text { as } k^2-3 k+1>0 \forall k \geq 3\end{aligned}$

$\begin{aligned} & =1+\sum_{k=3}^{100}\left(\frac{1}{(k-3)!}-\frac{1}{(k-1)!}\right) \\\\ & =1+\left(1-\frac{1}{2!}\right)+\left(\frac{1}{1!}-\frac{1}{3!}\right)+\left(\frac{1}{2!}-\frac{1}{4!}\right)+\ldots+ \\\\ & \qquad\left(\frac{1}{96!}-\frac{1}{98!}\right)+\left(\frac{1}{97!}-\frac{1}{99!}\right)\end{aligned}$

$\begin{aligned} & =3-\frac{1}{98!}-\frac{1}{99!}=3-\frac{9900}{100!}-\frac{100}{100!} \\\\ & =3-\frac{10000}{100!}=3-\frac{(100)^2}{100!} \\\\ & \therefore \frac{100^2}{100!}+\sum_{k=1}^{100}\left|\left(k^2-3 k+1\right) S_k\right|=3 \end{aligned}$