Another question on series t.t?

Question

I need to find the sum of this one.

12
Σ (2)^n-1
n = 1
So,
2^1-1 = 1 S(n) = t(1) (1 - r^n)
2^2-1 = 2 ------------------------
2^3-1 = 4 1 - r
2^4-1 = 8 S(12) =1(1 - 4096)
2^5-1 = 16 -------------------------
-1
S(12) = (1 - 4096)/-1 = 4095

I checked my work and this didn't work out to be the sum of them all ;/

Answer 1

That's a geometric series:

1 + 2 + 4 + 8 ...
n = 12, a = 1, r = 2

Sum = a*(1-r^n) / (1-r)
= 4095

UNLESS the question is really (2^n) - 1
Then the 2^n would constitute hte geometric series, and the -1 added 12 times will sum to -12.
So in this case the first term of the GP is 2 NOT 1, so the sum will be 2*4095 - 12 = 8178.

Depends on exactly what the question is.

Answer 2

It should work out. Make sure you have the right 2^11th power, which is 2048.