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## anonymous 5 years ago For the geometric series 1 + 3 + 9 + 27 + . . . , find the sum of the first 10 terms. does anyone know how to solve this?

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1. anonymous

each one is *3 so 1+3+9+27+81+243+729+2187+6561+19683 = 29524. the correct way to do this is using factorials (the exclamation point symbol) factorial of 3 over a range of 10

2. anonymous

Loss, that isn't help ... "sum it by writing out ever terms and adding them ' ... good job.

3. anonymous

I'll post the correct method:

4. anonymous

ok thanks

5. anonymous

Call the geometric series, Sn: $Sn = a + ar + ar^2 + ... + ar^{n-1}$ multiplying by r: $rSn = ar + ar^2 + ar^3 + ... + ar^{n}$ So we can say: $Sn - rSn = a - ar^n \implies Sn(1-r) = a(1-r^n)$ $\implies Sn = \frac{a(1-r^n)}{1-r}$ This is the general sum or a geometric series with starting term a and ratio r Can you see what your a and r are, and hence solve?

6. anonymous

* and n terms

7. anonymous

Oh, OK:( a = 1, r = 3, n = 2 $\implies Sn = \frac{1-3^{10}}{1-3} = 29524$ Oh, wow, it's the same as above.. except now I can solve any general geometric series. Taking the above further: $Sn = \frac{a(1-r^n)}{1-r}$ If |r| < 1, $r^n \rightarrow 0\ \text{as n } \rightarrow \infty$ So we have a general formula : $s_\infty = \frac{a}{1-r} \text{ for } |r| < 1$

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