Question

What is the sum of the geometric series E (-2)(-3)^n-1

Answer 1

Answer:

The sum is 40.

Step-by-step explanation:

Given,

$\sum_{n=1}^{4} (-2)(-3)^{n-1}$

We know that,

$\sum_{n=1}^{4} (-2)(-3)^{n-1}=(-2)(-3)^{1-1}+(-2)(-3)^{2-1}+(-2)(-3)^{3-1}+(-2)(-3)^{4-1}$

$=(-2)(-3)^{0}+(-2)(-3)^1+(-2)(-3)^2+(-2)(-3)^3$

$=-2\times 1 - 2\times - 3 - 2\times 9-2\times -27$

$=-2+6-18+54$

$=40$

Hence,

$\sum_{n=1}^{4} (-2)(-3)^{n-1}=40$

Answer 2

Sum of the geometric series: S=?
S=(-2)(-3)^(1-1)+(-2)(-3)^(2-1)+(-2)(-3)^(3-1)+(-2)(-3)^(4-1)
S=(-2)[(-3)^0+(-3)^1+(-3)^2+(-3)^3]
S=(-2)[1+(-3)+9+(-27)]
S=(-2)(1-3+9-27)
S=(-2)(-20)
S=40

Answer: Third option 40