Question

Find the sum of
$\frac{1}{6}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}$

Answer 1

We have,

$\frac{1}{6}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}$

Since,

$\begin{array}{c}\frac{1}{2\times 3}+\frac{1}{3\times 4}+\frac{1}{4\times 5}+\frac{1}{5\times 6}+\frac{1}{6\times 7}+\frac{1}{7\times 8}+\frac{1}{8\times 9}+\frac{1}{9\times 10}+\frac{1}{10\times 11}+\frac{1}{11\times 12}\\ =\frac{1}{6}+\left[\frac{\frac{1}{2}-\frac{1}{12}}{1}\right]\\ =\frac{1}{6}+\left[\frac{6-1}{12}\right]\\ =\frac{1}{6}+\frac{5}{12}\\ =\frac{2+5}{12}\\ =\frac{7}{12}\end{array}$

Hence, this is the answer.