Find the sum to `n` terms of the series: `1/(1+1^2+1^4)+1/(1+2^2+2^4)+1/(1+3^2+3^4)+`

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1 Answers

The nth term of the given series is `T_(n)=(n)/(1+n^(2)+n^(4))`
`:.`Sum of n terms `S_(n)=sumT_(n)=sum(n)/(1+n^(2)+n^(4))`
`=sum(n)/((1+n+n^(2))(1-n+n^(2)))`
`=(1)/(2)sum((1)/(1+n+n^(2))-(1)/(1-n+n^(2)))`
`=(1)/(2)((1)/(1-1+1)-(1)/(1+n+n^(2))) " " [" by property"]`
`=((n+n^(2)))/(2(1+n+n^(2)))=(n(n+1))/(2(n^(2)+n+1))`.

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