Question

$\sum _{r=1}^{50}[\frac{1}{49+r}-\frac{1}{2r(2r-1)}]=$

• A. $\frac{1}{50}$
• B. $\frac{1}{99}$
• C. $\frac{1}{100}$
• D. $\frac{1}{101}$

Answer 1

The correct option is C $\frac{1}{100}$

$\sum _{r=1}^{50}\frac{1}{2r(2r-1)}=\sum _{r=1}^{50}(\frac{1}{2r-1}-\frac{1}{2r})$

$=(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+........+\frac{1}{99}-\frac{1}{100})$

$=(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.........+\frac{1}{100})-2(\frac{1}{2}+\frac{1}{4}+............+\frac{1}{100})$

$=\sum {r=1}^{100}\frac{1}{r}-(1+\frac{1}{2}+.....+\frac{1}{50})$, i.e.,$\sum _{r=1}^{50}\frac{1}{r}=\sum _{r=51}^{100}\frac{1}{r}$

$\therefore E=\sum _{r=1}^{50}\frac{1}{49+r}-\sum _{r=51}^{100}\frac{1}{r}$

$=[\frac{1}{50}+\frac{1}{51}+...........+\frac{1}{99}]-[\frac{1}{51}+\frac{1}{52}+............+\frac{1}{100}]$

$=\frac{1}{50}-\frac{1}{100}=\frac{2}{100}-\frac{1}{100}=\frac{1}{100}\Rightarrow \left(C\right)$