Geometric series: sum of series?

If we had the following geometric series:

a + ar + ar^2 + ar^3 + ...

Then the sum of the terms from 0 to n (which we can call "S[n]" for sum to n) is:

a + ar + ar^2 + ar^3 + ... + ar^n = S[n]

But if we multiply both sides by r, and play around with the math, we get:

ar + ar^2 + ar^3 + ... + ar^(n+1) = rS[n]

a + ar + ar^2 + ar^3 + ... + ar^(n+1) = a + rS[n]

(a + ar + ar^2 + ar^3 + ... ar^n) + ar^(n+1) = a + rS[n]

S[n] + ar^(n+1) = a + rS[n]

S[n] (1 - r) + ar^(n+1) = a

S[n] = a(1 - r^(n+1)) / (1 - r)

This is the general formula for a geometric series

Now if we take n to infinity, and if r < 1, then r^(n+1) goes to zero. Also, if a=1, then we're left with 1 / (1-r). You can apply this to a geometric series that goes from 0 to infinity. So for the first example:

Σ [2 to ∞] (1/2)^n =

(Σ [0 to ∞] (1/2)^n) - 1 - 1/2 =

( 1 / (1 - 1/2)) - 1 - 1/2 =

( 1 / (1/2)) - 1 - 1/2 =

2 - 1 - 1/2 =

1/2

For the second problem, we have to do a little work to get something into the form of r^n:

Σ [0 to ∞] (2^n) / 5^(2n+1) =

Σ [0 to ∞] (2^n) / 5*5^(2n) =

(1/5) Σ [0 to ∞] (2^n) / 5^(2n) =

(1/5) Σ [0 to ∞] (2^n) / (5^2)^n =

(1/5) Σ [0 to ∞] (2^n) / (25)^n =

(1/5) Σ [0 to ∞] (2/25)^n =

(1/5) ( 1 / (1 - 2/25)) =

(1/5) ( 1 / (23/25)) =

(1/5) ( 25/23) =

5/23

You should get the idea now.

I'll just do the last one so you can get the gist of it.

I am assuming that you need brackets around the 6 and 2.

ie (6^n+2^n)/7^n = (6/7)^n + (2/7)^n

Consider this to be 2 geometric series

The first one is (6/7)^n

= 6/7 + (6/7)^2 + (6/7)^3 +...

= a / (1-r)

a = 6/7, r = 6/7

so the sum = (6/7) / (1/7) = 6

Similarly the other one has a sum of 2/5

So your final answer is 6 + 2/5 = 6.4

Use Sn = a/(1-r) where a = 1st term and r = common ratio.

e.g. for 1st one a = 1/4 and r = 1/2 so Sn = 1/4 /(1- 1/2) = 1/2

For the last one you will have to add TWO series together.