Find the sum : $\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}$ $+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}$.
(a)$\frac{7}{8}$
(b)$\frac{11}{12}$
(c)$\frac{15}{16}$
(d)$\frac{17}{18}$
Answer
Answer (as printed): B
Explanation
Given $\exp .=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)$ $$+\left(\frac{1}{4}-\frac{1}{5}\right)+\ldots . .+\left(\frac{1}{11}-\frac{1}{12}\right)=\left(1-\frac{1}{12}\right)=\frac{11}{12} .$$
Explanation as extracted from the printed page; notation may be imperfect.