# How do you find the sum given sum_(k=0)^4 1/(k^2+1)?

Oct 31, 2016

${\sum}_{k = 0}^{k = 4} \frac{1}{{k}^{2} + 1} = \frac{158}{85}$

#### Explanation:

${\sum}_{k = 0}^{k = 4} \frac{1}{{k}^{2} + 1} = \frac{1}{{0}^{2} + 1} + \frac{1}{{1}^{2} + 1} + \frac{1}{{2}^{2} + 1} + \frac{1}{{3}^{2} + 1} + \frac{1}{{4}^{2} + 1}$
$\therefore {\sum}_{k = 0}^{k = 4} \frac{1}{{k}^{2} + 1} = \frac{1}{0 + 1} + \frac{1}{1 + 1} + \frac{1}{4 + 1} + \frac{1}{9 + 1} + \frac{1}{16 + 1}$
$\therefore {\sum}_{k = 0}^{k = 4} \frac{1}{{k}^{2} + 1} = \frac{1}{1} + \frac{1}{2} + \frac{1}{5} + \frac{1}{10} + \frac{1}{17}$
$\therefore {\sum}_{k = 0}^{k = 4} \frac{1}{{k}^{2} + 1} = \frac{158}{85}$