Question

What is the sum of the geometric series?

Answer 1

The sum is C. 40.

In most other cases you would do it with a formula, but because there are only 4 terms, we can do this by hand.

(-2)(-3)^1-1 = (-2)(1) = -2
(-2)(-3)^2-1 = (-2)(-3) = 6
(-2)(-3)^3-1 = (-2)(9) = -18
(-2)(-3)^4-1 = (-2)(-27) = 54

- 2 + 6 - 18 + 54 = 40.

Answer 2

Option 3 rd is correct

sum of the given geometric series is, 40

Explanation:

The sum of the finite geometric series is given by:

`S_n=(a_1(1-r^n))/(1-r)`
`r\\eq 1` ....[1]

Given that:

`\sum_(n=1)^(4) (-2)(-3)^(n-1)`

We have to find the
`S_(4)` for the series:

`(-2)(-3)^0+(-2)(-3)^1+(-2)(-3)^2+(-2)(-3)^3`

`-2+6-18+54`

here,
`a_1 = -2` and common ratio(r) = -3

Substitute these in [1] we have;

`S_4=(-2 \cdot (1-(-3)^(4)))/(1-(-3))`

⇒
`S_4=(-2 \cdot (1-(81)))/(1+3)`

⇒
`S_4=(-2 \cdot -80)/(4)`

Simplify:

`S_4 =(160)/(4) = 40`

therefore, the sum of the given geometric series is, 40