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 $\displaystyle\sum_{ k =1}^{100}\left(\frac{1}{ S _{ k }}\right)$ 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+\ldots \ldots+ k =\frac{ k ( k +1)}{2} $
$\therefore \displaystyle\sum_{ k =1}^{100}\left(\frac{1}{ S _{ k }}\right)=\displaystyle\sum_{ k =1}^{100}\left(\frac{2}{ k ( k +1)}\right)=2 \displaystyle\sum_{ k =1}^{100}\left(\frac{1}{ k }-\frac{1}{ k +1}\right)$
$ =2\left[\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\ldots \ldots+\left(\frac{1}{100}-\frac{1}{101}\right)\right]=2\left[1-\frac{1}{101}\right]=\frac{200}{101}$