Try this series

ok i have waited long enough for a solution, now posting the solution

$\frac{4}{1\times 2}+\frac{12}{2\times 5}+\frac{20}{5\times 10}+\frac{28}{10\times 17}+\frac{36}{17\times 26}$

Apply method of difference on denominator to find the general term.

The general term $T_{r}=\huge{\frac{4(2r-1)}{(r^2+2-2r)(r^2+1)}}$

Number of terms $5$

To find $\displaystyle \sum_{r=1}^{5} T_{r}$

$\displaystyle \sum_{r=1}^{5} 4(\frac{2r-1}{((r-1)^{2}+1)(r^2+1)})$

$\displaystyle \sum_{r=1}^{5} 4(\frac{r^2+1-((r-1)^{2}+1)}{((r-1)^{2}+1)(r^2+1)})$

$\displaystyle \sum_{r=1}^{5} 4 (\frac{1}{(r-1)^{2}+1}-\frac{1}{r^2+1})$

Apply $V_{n}$ method or u can simply see that except the first and the last term , all the terms get cancelled.

$4(1-\frac{1}{5^2+1})$

$4(\frac{25}{26})$

$\large{\boxed{\frac{50}{13}}}$