Finding a of n from Sn partial sum

[freshman2013]

Homework Statement

suppose that the partial sum of the series (sigma)n=1,infinity a_n is given by the partial sum S_n = (-2n+9)/(-6n+15). Find an expression for a_n when n>1

Homework Equations

S_n= (-2n+9)/(6n+15

The Attempt at a Solution

So I attempted to subtract S(n-1) from S(n) to get each term for a_n and got the following terms
a₂=8/9
3=-8/3
4=8/9
5=8/45
6=8/105
7=8/189
8=8/297
How am I supposed to come up with a generalized expression from these terms, or am I wrong from the first step of doing S(n)-S(n-1) to get those terms for a_n? The only pattern I can recognize is that after a₄, the difference of the denominators increase by 24 from one term to the next.

Homework Statement

Homework Equations

The Attempt at a Solution

 

[Dick]

Homework Statement

suppose that the partial sum of the series (sigma)n=1,infinity a_n is given by the partial sum S_n = (-2n+9)/(6n+15). Find an expression for a_n when n>1

Homework Equations

S_n= (-2n+9)/(6n+15

The Attempt at a Solution

So I attempted to subtract S(n-1) from S(n) to get each term for a_n and got the following terms
a₂=8/9
3=-8/3
4=8/9
5=8/45
6=8/105
7=8/189
8=8/297
How am I supposed to come up with a generalized expression from these terms, or am I wrong from the first step of doing S(n)-S(n-1) to get those terms for a_n? The only pattern I can recognize is that after a₄, the difference of the denominators increase by 24 from one term to the next.

Homework Statement

Homework Equations

The Attempt at a Solution

$S_{n}-S_{n-1}=a_n$. So take your expression for $S_n$ and subtract the same expression with n-1 substituted for n. Putting numbers in isn't the way to do it.