Question

(3^2+1)/(3^2-1)+(5^2+1)/(5^2-1)+(7^2+1)/(7^2-1)+......+(17^2+1)/(17^2-1)

计算题：(3^2+1)/(3^2-1)+(5^2+1)/(5^2-1)+(7^2+1)/(7^2-1)+......+(17^2+1)/(17^2-1)
请尽快回答！！！！！！！

Answer 1

（3^2+1）/(3^2-1) = 1+2/((3+1)(3-1)) = 1+1/2-1/4
（5^2+1）/(5^2-1) = 1+2/((5+1)(5-1)) = 1+1/4-1/6
……
（17^2+1）/(17^2-1) = 1+2/((17+1)(17-1)) = 1+1/16-1/18

所以
原式 = 8 + 1/2-1/4+1/4-1/6+……+1/16-1/18 =8+1/2-1/18 = 8+4/9

Answer 2

(3^2+1)/(3^2-1)+(5^2+1)/(5^2-1)+(7^2+1)/(7^2-1)+......+(17^2+1)/(17^2-1)

=1+2/(3^2-1)+1+2/(5^2-1)+1+2/(7^2-1)+......+1+2/(17^2-1)
=8+2/(3+1)(3-1)+2/(5+1)(5-1)+2/(7+1)(7-1)+......++2/(17+1)(17-1)
=8+2/(2×4)+2/(4×6)+2/(6×8)+……+2/(16×18)
=8+(1/2-1/4)+(1/4-1/6)+(1/6-1/8)+……+(1/16-1/18)
=8+1/2-1/18
=8+4/9
=76/9

Answer 3

2

Answer 4

186999