Question

For the geometric series –8, 4, –2, . . . find the following sums:

a. s3

b. s6

c. s25

d. S, the sum of the series

Answer 1

geometric series:

`(4)/(-8) = - (1)/(2)`

`(-2)/(4) = - (1)/(2)`

`\implies r = - (1)/(2) \ \ and \ \ a_(1) = -8`

`\boxed{S_(n) = \frac{a_(1)\cdot (1 - {r}^(n)) }{1 - r}}`

a. n = 3

`S_(3) = - 8 + 4 - 2 = 4 - 10 = \bf -6\\`

b. n = 6

`S_(6) = \frac{(-8)\cdot \Big(1 - { \Big(- (1)/(2)\Big)}^(6)\Big) }{1 - \Big( - (1)/(2)\Big)} = ((-8)\cdot \Big(1 - (1)/(2^(6))\Big))/(1 + (1)/(2)) = (-8 + (1)/(2^(3)))/((3)/(2)) =\\`

`= (-64+1)/(2^(3)) \cdot (2)/(3) = -(63)/(2^(2)(3)) = \bf -(21)/(4)`

c, n = 25

`S_(25) = \frac{(-8)\cdot \Big(1 - { \Big(- (1)/(2)\Big)}^(25)\Big) }{1 - \Big( - (1)/(2)\Big)} = ((-8)\cdot \Big(1 + (1)/(2^(25))\Big))/(1 + (1)/(2)) = - (8 + (1)/(2^(22)))/((3)/(2)) =\\`

`= - (8 + (1)/(2^(22)))/((3)/(2)) = - (1 + 2^(25))/(2^(22)) \cdot (2)/(3) = \bf - (1 + 2^(25))/((3)2^(21))\\`