Question

Find the sum of the following series. Round to the nearest hundredth if necessary,

3 + 6 + 12+ ... +49152
Sum of a finite geometric series:
Q] - a
Sn =
1-T

Answer 1

`sum = \frac{a( {r}^(n - 1) )}{r - 1} \\ : but \: l = a( {r}^(n - 1) ) \\ 49152 = 3( {2}^(n - 1) ) \\ 16384 = {2}^(n - 1) \\ {2}^(n) = 32768 \\ {2}^(n) = {2}^(15) \\ n = 15 \\ \therefore \: sum = \frac{3(2 {}^(15 - 1)) }{15 - 1} \\ = (49152)/(14) \\ = 3510.9`