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)

Question

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)

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.

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

Related Questions

If, in a hydrogen atom, radius of nth Bohr orbit is `r_(n)` frequency of revolution of electron in nth orbit is `f_(n)` and area enclosed by the nth o

2 Answers 4 Views

Find the sum of n terms of the series whose nth terms is (i) `n(n-1)(n+1)`.

2 Answers 4 Views

Find the sum of n terms of the series whose nth terms is (ii) `n^(2)+3^(n)`.

2 Answers 4 Views

Statement 1 The sum of first n terms of the series `1^(2)-2^(2)+3^(2)-4^(2)_5^(2)-"……"` can be `=+-(n(n+1))/(2)`. Statement 2 Sum of first n narural n

2 Answers 4 Views

Consider an AP with a as the first term and d is the common difference such that `S_(n)` denotes the sum to n terms and `a_(n)` denotes the nth term o

2 Answers 5 Views

If the sum of the first ten terms of an `A.P` is four times the sum of its first five terms, the ratio of the first term to the common difference is:

2 Answers 4 Views

For the series, `S=1+1/((1+3))(1+2)^2+1/((1+3+5))(1+2+3)^2+1/((1+3+5+7))(1+2+3+4)^2` +... 7th term is 16 7th term is 18 Sum of first 10 terms is `(505

2 Answers 4 Views

`S_(n)` be the sum of n terms of the series `(8)/(5)+(16)/(65)+(24)/(325)+"......"` The seveth term of the series is

2 Answers 4 Views

Satement 1 If the nth termn of a series is `2n^(3)+3n^(2)-4`, then the second order differences must be an AP. Statement 2 If nth term of a series is

2 Answers 4 Views

If `s_1,s_2,s_3` denote the sum of n terms of 3 arithmetic series whose first terms are unity and their common difference are in H.P., Prove that `n=(

2 Answers 4 Views

Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

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.