Question

sum (k^2 + 1)/(k^2 - 1) k=2 to k=90
How?

Answer 1

(k^2+1)/(k^2-1) = 1 + 1/(k-1) - 1/(k+1)
now the sum becomes
(1 + 1/1 - 1/3)+(1 + 1/3 - 1/5) + ... + (1 + 1/89 - 1/91)
There are 89 sets of terms, so the result is
89 + 1/1 - 1/91

Answer 2

What does "how" mean?
I will assume you want

∑ (k^2+1)/(k^2-1) for k from 2 to 90 ??

found this to be
= (2n+1)(n^2 + n -2)/((2n(n+1)) , for n ≥ 2 , using Mathemagics

so if n = 90
sum = 181(8100 + 90 - 2)/(180(91)) = 1482028/16380
= 370507/4095

testing: if n = 6, you have
5/3 + 10/8 + 17/15 + 26/24 + 37/35 = 130/21
using my formula: sum = 13(36+6-2)/(12*7)
= 520/84
= 130/21

Answer 3

Another look at oobleck's :
He probably got
(k^2 + 1)/(k^2 - 1) = 1 + 1/(k-1) - 1/(k+1) using partial fractions

So our series is, starting with k = 2
= (1 + 1/1 - 1/3) + (1 + 1/2 - 1/4) + (1 + 1/3 - 1/5) +...+(1+1/89 - 1/91)
= 89 + (1/1 + 1/2 + 1/3 +...+ 1/87 + 1/88 + 1/89) - (1/3 + 1/4 + ... + 1/90+1/91)
this telescopes to
89 +1/1 + 1/2 - 1/90 - 1/91
= 90 + 1957/4095
= 370507/4095 , the same as when I used the "formula

Answer 4

good catch. I missed all the even ones.
My Bad.