SOLUTION: Compute the sum 1 + 1/3 + 2/3 + 2/9 + 4/9 + 4/27 + 8/27 + 8/81... Thanks!

Question 1036461: Compute the sum
1 + 1/3 + 2/3 + 2/9 + 4/9 + 4/27 + 8/27 + 8/81...

Thanks!
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
it looks like you have 2 geometric sequences going here.

the first geometric sequence is composed of all odd number terms.

1 + 2/3 + 4/9 + 8/27 ...

the common ratio would be 2/3.

1 * 2/3 = 2/3 * 2/3 = 4/9 * 2/3 = 8/27, ...

the second geometric sequence is composed of all even number terms.

1/3 + 2/9 + 4/27 + ...

the common ratio is again 2/3.

1/3 * 2/3 = 2/9 * 2/3 = 4/27, ...

if you take the sum of both geometric series and add them together, you should get the sum of the combined series.

you can test this out by taking the sum of each series to the 3d term.

this will show you that the formula works.

once you are assured of that, you can then take the sum to infinity.

the sum of a geometric series is Sn = A1 * (1 - r^n) divided by (1 - r).

in the first geometric series, we would get S3 = 1 * (1 - (2/3)^3) divided by (1 - 2/3) which would be equal to 1 * 2.11111

add 1 and 2/3 and 4/9 together to get 2.111...

the formula works for the first geometric series.

in the second geometric series, we would get S3 = 1/3 * (1 - 2/3)^3) divided by (1 - 2/3) which would be equal to .7037037037.

add 1/3 and 2/9 and 4/27 together to get .7037037037.

the formula works for the second geometric series.

the first 3 terms of the first geometric series added to the first 3 terms of the second geometric series gets you.

1 + 1/3 + 2/3 + 2/9 + 4/9 + 4/27

add these together and you get 2.814814815

add 2.1111111111 and .7037037037. together and you get 2.814814815.

the formula works for the combined series.

now that you know it works, you can solve for the infinite series.

the formula for the infinite series is S∞ = A1 * 1 / (1-r)

since 1/(1-r) = 1/(1-2/3) = 1/(1/3) = 3, you would get 1 * 3 + 1/3 * 3 = 4.

the sum of the infinite series is 4.

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