Question

What is the sum of the geometric series?

Answer 1

Answer:

Option 3 rd is correct

sum of the given geometric series is, 40

Step-by-step explanation:

The sum of the finite geometric series is given by:

$S_n=\frac{a_1(1-r^n)}{1-r}$ $r\neq 1$        ....[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;

$S_4=\frac{-2 \cdot (1-(-3)^{4})}{1-(-3)}$

⇒ $S_4=\frac{-2 \cdot (1-(81))}{1+3}$

⇒ $S_4=\frac{-2 \cdot -80}{4}$

Simplify:

$S_4 =\frac{160}{4} = 40$

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

Answer 2

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.