# How do you find the sum of #Sigma 1/(j^2-3)# where j is [3,5]?

##### 1 Answer

Aug 17, 2017

#### Answer:

# 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#