How do you find the sum of #Sigma 1/(k^2+1)# where k is [0,3]?
1 Answer
Aug 12, 2017
# 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 #
