Question

Let $S_k,k=1,2,\dots ,100$, 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 $\frac{{100}^2}{100!}+\sum _{k=2}^{100}|(k^2-3k+1)S_k|$ is :

• A. $3$
• B. $1$
• C. $2$
• D. $4$

Answer 1

Answer: (A)

$3$

$S_k=\frac{\frac{k-1}{k!}}{1-\frac{1}{k}}=\frac{1}{(k-1)!}$, for $k>1$

$\sum _{k=2}^{100}|(k^2-3k+1)\frac{1}{(k-1)!}|$
$=\sum _{k=2}^{100}\left|\frac{(k-1{)}^2-k}{(k-1)!}\right|$
$=\sum _{k=2}^{100}|\frac{k-1}{(k-2)!}-\frac{k}{(k-1)!}|$

$=|\frac{1}{0!}-\frac{2}{1!}|+|\frac{2}{1!}-\frac{3}{2!}|+|\frac{3}{2!}-\frac{4}{3!}|+\cdots$
$=\frac{2}{1!}-\frac{1}{0!}+\frac{2}{1!}-\frac{3}{2!}+\frac{3}{2!}-\frac{4}{3!}+\cdots +\frac{99}{98!}-\frac{100}{99!}$
$=1+\frac{2}{1!}-\frac{100}{99!}$
$=3-\frac{100}{99!}$

Hence the value of $\frac{{100}^2}{100!}+\sum _{k=1}^{100}|(k^2-3k+1)S_k|$
$=\frac{100}{99!}+3-\frac{100}{99!}$

$=3$