A-level exponential and logarithmic functions question?

How many terms of the geometric series 1+(1/2)+(1/4)+(1/8 )+.... must be taken for the sum to differ from 2 by less than 10^-8?

The answer is penis.

...Thanks

...Thanks

Surely you wouldn't expect any other answer since 99% of A Level students are dreading the few hours to come...

Penis.

a = 1, r = 1/2. Solve 2 - a(1 - r^n)/(1 - r) < 10^-8 and note the inequality sign must flip.

sum of geometric series = a(1-r^n)/(1-r)

a= 1 r= 1/2

sum = 1(1-0.5^n)/(0.5) = 1.99999999
1-0.5^n = 1.99999999 x 0.5
1 - (1.99999999 x 0.5) = 0.5^n
0.000000005 = 0.5^n
log 0.000000005 = n log 0.5
n = 27.57.....

when n = 27, sum = 1.999999985
when n = 28, sum = 1.999999993

so 28 terms