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

Answer:

It's the first choice.

Step-by-step 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

Answer:

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:

        $\sum_{n=0}^{\infty} 2(-3)^n$

Step-by-step explanation:

The alternating series is given by:

                       $2-6+18-54+........$

The series could also be written in the form:

$=2+(2\times (-3))+(2\times (-3)\times (-3))+(2\times (-3)\times (-3)\times (-3))+....\\\\i.e.\\\\=2\times (-3)^0+2\times (-3)^1+2\times (-3)^2+2\times (-3)^3+.....\\\\i.e.\\\\=\sum_{n=0}^{\infty} 2(-3)^n$