Question

Write the sum using summation notation, assuming the suggested pattern continues. 2 - 6 + 18 - 54 + ... summation of two times negative three to the power of n from n equals zero to infinity summation of two times three to the power of n from n equals zero to infinity summation of two times three to the power of the quantity n plus one from n equals zero to infinity summation of two times negative three to the power of the quantity n minus one from n equals zero to infinity

Answer 1

It's the first choice.

Explanation:

The common ratio is -6/2 = 18/-6 = -3.

2*(-3)^0 = 2*1 = 2.

2*(-3)^1 = -6

2*(-3)^2 = 18

2*(-3)^3 = -54.

So in summation notation is

∞

∑ 2(-3)^n

n=0

Answer 2

The sum using summation notation is given by:

Summation of two times negative three to the power of n from n equals zero to infinity.

i.e. numerically it is given by:

`\sum_(n=0)^(\infty) 2(-3)^n`

Explanation:

The alternating series is given by:

`2-6+18-54+........`

The series could also be written in the form:

`=2+(2* (-3))+(2* (-3)* (-3))+(2* (-3)* (-3)* (-3))+....\\\\i.e.\\\\=2* (-3)^0+2* (-3)^1+2* (-3)^2+2* (-3)^3+.....\\\\i.e.\\\\=\sum_(n=0)^(\infty) 2(-3)^n`