Question

Let $S_k$ k = 1, 2, ....., 100 denote the sum of the infinite geometric series whose first term is $\frac{k-1}{k!}$ and the common ratio is $\frac{1}{k}$. Then the value of $\frac{{100}^2}{100!}\sum _{k=1}^{100}\left|(k^2-3k+1)S_k\right|$ is

• A. $3$
• B. $4$
• C. $5$
• D. $7$

Answer 1

The correct option is A $3$

Infinite Sum of GP $S=\frac{a}{1-r}=\frac{(k-1)}{k!(1-1/k)}=\frac{1}{(k-1)!}$

$\frac{{100}^2}{100!}+{\sum }_{k=1}^{100}\left|\right(k^2-3k+1\left)S_k\right|\frac{{100}^2}{100!}+{\sum }_{k=1}^{100}\left[\right({(k-1)}^2-k\frac{1}{(k-1)!=}\left)\right]=\frac{{100}^2}{100!}+{\sum }_{k=2}^{100}|\frac{(k-1)}{(k-2)!}-\frac{k}{(k-1)!}|\frac{{100}^2}{100!}+\frac{2}{1!}-\frac{1}{0!}+\frac{2}{1!}-\frac{3}{2!}+\frac{3}{2!}+.....+\frac{99}{98!}-\frac{100}{99!}\frac{{100}^2}{100!}+2-1+2-dfrac\{100\times 100\}\{99!\times 100\}=3$