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.

Explanation

Since, `t_(n)=(n^(5)+n^(3))/(n^(4)+n^(2)+1)`
`=n-(n)/(n^(4)+n^(2)+1)`
`=n+(1)/(2(n^(2)+n+1))-(1)/(2(n^(2)-n+1))`
Sum of n terms `S_(n)=sumt_(n)`
`=sumn+(1)/(2){sum((1)/(n^(2)+n+1)-(1)/(n^(2)-n+1))}`
`=(n(n+1))/(2)+(1)/(2)((1)/(n^(2)+n+1)-1)" " [" by property "]`
`(n^(2))/(2)+(n)/(2)=-(1)/(2)+(1)/((nsqrt(2)+(1)/(sqrt(2)))^(2)+(3)/(2))`
but given `S_(n)=a_(n)^(2)+a(1)/(b_(n)^(2)+b)`
On comparing, we get
`a_(n)=(n)/(sqrt(2))+(1)/(2sqrt(2)),a=-(5)/(8),b_(n)=(nsqrt(2)+(1)/(sqrt(2))),b=(3)/(2)`
`:. (b_(n))/(a_(n))=2` , which is constant.