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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?

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4 Answers

up vote 1 down vote accepted

$S_n-S_{n-1}=a_n$, 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.

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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 :*(

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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.

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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 $$\frac{a_{n+1}}{a_1}=\frac{3(2^{n+1}+1)}{3(2^n+1)}=\frac{2^{n+1}+1}{2^n+1}$$

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what he has listed are the partial sums, not the sequence itself, so those fractions don't represent ratios of successive terms. –  Robert Mastragostino Jun 3 '12 at 23:11
    
Indeed, you have $S_{n+1}/S_n$. –  anon Jun 3 '12 at 23:12
    
That's interesting. I assumed it was geometric because that was what we were learning. Thank you. –  InQ Jun 3 '12 at 23:28
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