Question

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

Answer 1

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

Answer 2

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

Answer 3

I'll post the correct method:

Answer 4

ok thanks

Answer 5

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?

Answer 6

* and n terms

Answer 7

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 :
\[s_\infty = \frac{a}{1-r} \text{ for } |r| < 1\]