求和 1/(1+1的平方+1的4次方)+2(1+2的平方+2的4次方).+100/(1+100的平方+100的

2/(1+2的平方+2的4次方)应该是少了一个除号吧.
n/(1+n^2+n^4)
=n/[(n^2+1)^2 -n^2]
=n/(n^2+n+1)(n^2-n+1)
=0.5*[1/(n^2-n+1) -1/(n^2+n+1)]
=0.5*[1/(n^2-n+1) -1/[(n+1)^2-(n+1)+1]
原式=0.5*[1/(1^2-1+1) -1/(2^2-2+1) +1/(2^2-2+1)-1/(3^2-3+1)+.+1/(100^2-100+1)-1/(101^2-101+1)]
=0.5*[1/(1^2-1+1) -1/(101^2-101+1)]
=5050/10101

an=1/(1+n^2+n^4)
∵a³-b³=(a-b)(a²+ab+b²)
∴a²+ab+b²=(a³-b³)/(a-b)
∴1/(1+n^2+n^4)=(n²-1)/(n^6-1)