Find nth term given the equation of a series Given $S_n=3(2^n+1)$, find the fourth term.
I believe I can find the first term, $a$, assuming n=1 is the beginning. $3(2^1+1)=9$
To find the fourth term, I have tried to do it by subtracting $S_4$ and $S_3$. This gives me 24.
Is this mathematically sound? When I try to determine the common ratio from this, namely $24=ar^3$, $r=1.386722549$. However, when I try to find another term, say $t_3$, $S_3-S_2 \neq ar^2$. Therefore, I must be doing something wrong here; specifically, finding the common ratio from only one equation is probably incorrect.
So:
1) Is 24 the third term?
2) If so, how do I find the common ratio from this?
 A: The elements of any sequence $\,\{a_n\}\,$ of which its partials sums $\,\{S_n\}\,$ are given can be easily evaluated as you did $$a_n=S_n-S_{n-1}$$ Your problem is that you're wrongly assuming the sequence is *geometric...and it isn't, as you can readily check by evaluating $$\frac{a_{n+1}}{a_1}=\frac{3(2^{n+1}+1)}{3(2^n+1)}=\frac{2^{n+1}+1}{2^n+1}$$
A: $S_n-S_{n-1}=a_n$, right? You get $S_n$ by adding $a_n$ to $S_{n-1}$. So then:
$$a_n=S_n-S_{n-1}=3(2^n+1)-3(2^{n-1}+1)=3(2^n-2^{n-1})=3\cdot2^{n-1}$$
So
$$a_4=3\cdot 2^3=24$$
Yes the method is perfectly fine. The issue you had stems from finding an incorrect value of $a$. Use the method of subtracting sums that you used in the first half of the question:
$$a=S_1-S_0=9-6=3$$ which is the correct value.
A: As mentioned above, there is no need to assume that the sequence has an arithmetic or geometric form, and therefore there may not be a common difference or common ratio.
You correctly deduced the 1st term, and could find the nth term iteratively, if necessary. Using the same method that you did for $a_1$, calculate $S_2, S_3,$ and $S_4$ and use your formula 
$a_n=S_n-S_{n-1}$ to find the rest.
A: You are correct the partial sum $S_n=\sum_{k=1}^n a_k$ has $n$-th term  $a_n=S_{n}-S_{n-1}$ because $S_n=S_{n-1}+a_n$ for all $n>0$.
Your fourth term is correct at 24
$S_4-S_3=3(17)-3(9)=3(8)=24$
Now you are assuming it is a geometric sum. That is $a_n=ar^n$ for some choice of $r$ and $a$ that work for all $n$. Note this usually doesn't work. In this case it actually doesn't work...
edited to fix sum isn't geometric :*(
