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

1 Answer
Oct 31, 2016

sum_(k=0)^(k=4) 1/(k^2+1) = 158/85

Explanation:

sum_(k=0)^(k=4) 1/(k^2+1) = 1/(0^2+1) + 1/(1^2+1) + 1/(2^2+1) + 1/(3^2+1) + 1/(4^2+1)
:. sum_(k=0)^(k=4) 1/(k^2+1) = 1/(0+1) + 1/(1+1) + 1/(4+1) + 1/(9+1) + 1/(16+1)
:. sum_(k=0)^(k=4) 1/(k^2+1) = 1/1 + 1/2 + 1/5 + 1/10 + 1/17
:. sum_(k=0)^(k=4) 1/(k^2+1) = 158/85