Question

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

Answer 1

The sum is 40.

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* 1 - 2* - 3 - 2* 9-2* -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