Question

What is the sum of the geometric sequence 1, 3, 9, … if there are 12 terms?

Answer 1

r = 3

a = 1 r>1

Answer 2

sum= a{r^n - 1}/r-1

Answer 3

if we construct a general setup of a geo seq:

a1 = a1
a2 = a1*r
a3 = a1 * r * r
a4 = a1 * r * r * r; of a1*r^3

a5 = a1* r^6
a6 = a1*r^5

an = a1*r^(n-1)

if we add up all these terms we get:

\[S = a1*r^0 + a1*r^1 + a1*r^2 + ... + a1*r^{n-3} + a1*r^{n-2}+ a1*r^{n-1}\]

factor out the a1a

\[a1*S = r^0 + r^1 + r^2 + ... + r^{n-3} + r^{n-2}+ r^{n-1}\]

the trick left is to multiply S by r to get

\[S*r = r^1 + r^2 + ... + r^{n-3} + r^{n-2}+ r^{n-1}+r^{n}\]

we then subtract one from the other

\[\hspace{5em} \ \ S = r^0 + r^1 + r^2 + ... + r^{n-3} + r^{n-2}+ r^{n-1}\]\[\hspace{4em}-(S*r) = -r^1 - r^2 - ... - r^{n-3} - r^{n-2}- r^{n-1}-r^{n}\]-----------------------------------------------------------\[S-(S*r) = r^0 \hspace{16em} - r^{n}\]

that was fun ....

now lets see; \[S -(S)*r=S(1-r)\]

divide both sides by (1-r)\[S=\frac{1-r^n}{1-r}\]
\[S=\frac{1-r^n}{1-r}\]

distribute the a1 back thru the summation to get\[S=a1\frac{1-r^n}{1-r}\]

Answer 4

@amistre64 that is only valid when r<1

Answer 5

here r>1

Answer 6

since we are not taking a limit to infinity; the summation still holds

it is when we want to know if the sum of an infinite geometric sequence converges that we have to consider the magnitude of r

Answer 7

http://www.wolframalpha.com/input/?i=%281-%283%29%5E12%29%2F%281-3%29

if i factor out the negative from the bottom and push it thru the top we get your version

http://www.wolframalpha.com/input/?i=%28%283%29%5E12-1%29%2F%283-1%29

which is identical ...

Answer 8

@Yahoo!