How do you find the sum of #Sigma 1/(j^2-3)# where j is [3,5]?
1 Answer
Aug 17, 2017
# sum_(j=3)^5 = 124/429#
Explanation:
With so few term involved, the easiest approach is just write out the terms and compute the sum:
# sum_(j=3)^5 1/(j^2-3) = 1/(3^2-3) + 1/(4^2-3) + 1/(5^2-3)#
# " " = 1/(9-3) + 1/(16-3) + 1/(25-3)#
# " " = 1/(6) + 1/(13) + 1/(22)#
# " " = 124/429#
