Question

Let $S_k$ be the sum from the first term to the $k^{\text{th }}$ term of the arithmetic sequence with the first term unity and common difference 1 . Then $\sum _{k=1}^{100}\left(\frac{1}{S_k}\right)$ equals

• A. $\frac{99}{50}$
• B. $\frac{200}{101}$
• C. $\frac{19}{10}$
• D. $\frac{100}{51}$

Answer 1

Answer: (B)

$S_k=1+2+3+\dots \dots +k=\frac{k(k+1)}{2}$
$\therefore \sum _{k=1}^{100}\left(\frac{1}{S_k}\right)=\sum _{k=1}^{100}\left(\frac{2}{k(k+1)}\right)=2\sum _{k=1}^{100}\left(\frac{1}{k}-\frac{1}{k+1}\right)$
$=2\left[\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\dots \dots +\left(\frac{1}{100}-\frac{1}{101}\right)\right]=2\left[1-\frac{1}{101}\right]=\frac{200}{101}$