1/(1+1^2+1^4)+2/(1+2^2+2^4)+3/(1+3^2+3^4)+……100/(1+100^2+100^4)

Question

1/(1+1^2+1^4)+2/(1+2^2+2^4)+3/(1+3^2+3^4)+……100/(1+100^2+100^4)

Answer 1

1/(1+1^2+1^4)分母是公比为1的等比数列
2/(1+2^2+2^4分母)是公比为2^2的等差数列
.
n/(1+n^2+n^4)分母是公比为n^2的等比数列
令分母为S
S=1*(1-n^6)/(1-n^2)=(1-n^3)(1+n^3)/(1+n)(1-n)=(1-n)(n^2+n+1)(1+n)(n^2-n+1)/(1-n)(1+n)=(n^2-n+1)(n^2+n+1)（公比不为1的等比数列求和公式,该数列只有三项）
分子除以分母即n/s=n/(n^2-n+1)(n^2+n+1)=(1/2)*[1/(n^2-n+1)-1/(n^2+n+1)](通项)
1/(1+1^2+1^4)+2/(1+2^2+2^4)+3/(1+3^2+3^4)+……100/(1+100^2+100^4)
=(1/2)*{(1-1/3)+(1/3-1/7)+(1/7-1/13)+.[1/(100^2-100+1)-1/(100^2+100+1)]}
=(1/2)*[1-1/(100^2+100+1)]
=(1/2)*(1-1/10101)
=(1/2)*(10100/10101)
=5050/10101