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The nth term of a series is given by t(n...

The nth term of a series is given by `t_(n)=(n^(5)+n^(3))/(n^(4)+n^(2)+1)` and if sum of its n terms can be expressed as `S_(n)=a_(n)^(2)+a+(1)/(b_(n)^(2)+b)` where `a_(n)` and `b_(n)` are the nth terms of some arithmetic progressions and a, b are some constants, prove that `(b_(n))/(a_(n))` is a costant.

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