Question

How do you find the sum of `Sigma 1/(k^2+1)` where k is [0,3]?

Answer 1

`sum_(k=0)^(3) 1/(k^2+1) = 9 /5`

Explanation:

We seek:

`sum_(k=0)^(3) 1/(k^2+1) = 1/(0^2+1) + 1/(1^2+1) + 1/(2^2+1) + 1/(3^2+1)`
`" " = 1/(0+1) + 1/(1+1) + 1/(4+1) + 1/(9+1)`
`" " = 1/1 + 1/2 + 1/5 + 1/10`
`" " = 9 /5`