Question

Sum of $\frac{1}{2.7}+\frac{1}{7.12}+\frac{1}{12.17}+\frac{1}{17.22}+...$ to n terms.

Answer 1

$\frac{1}{2.7}+\frac{1}{7.12}+\frac{1}{12.17}+.....nterms$

$=\frac{1}{2.7}+\frac{1}{7.12}=\frac{1}{12.17}+....+\frac{1}{(2+5(n-1\left)\right)(7+5(n-1)}$

$=\frac{1}{5}(\frac{5}{2.7}+\frac{5}{7.12}+.....+\frac{5}{(2+5(n-1\left)\right)(7+5(n-1\left)\right)})$

$=\frac{1}{5}[\frac{7-2}{2.7}+\frac{12-17}{7.12}+....+\frac{(7+5(n-1\left)\right)-(2+5(n-1\left)\right)}{2+5(n-1)\left)\right(7+5(n-1))}]$

$=\frac{1}{5}[\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+....+\frac{1}{2+5(n-1)}-\frac{1}{7+5(n-1)}]$

$=\frac{1}{5}\left[\frac{7+5(n-1)-2}{2.(7+5(n-1\left)\right)}\right]$

$=\frac{1}{5}\left[\frac{5+5(n-1)}{2.(7+(5(n-1))}\right]$

$=\frac{n}{14+10(n-1)}$

$=\frac{n}{10n+4}$