how do i find the sum of a geometric sequence

Hi CPhill,

Your correct answer is: $$2(3^5 -1 ) \textcolor[rgb]{1,0,0}{/} (3-1) = 242$$

It is equal to: $$s_n=\frac{a*(r^n-1)}{r-1}$$

here: $$s_5=\frac{2*(3^5-1)}{3-1}=242$$

Many Greetings

Heureka

a(r^n -1) / (r-1)

Where:

a is the first term

r is the ratio between a successive term and a previous term

n is the number of terms in the sum

For instance

2 6 18 54 162.....     2 = a    r = 3  n= 5

2(3^5 -1 ) (3-1) = 242

Thanks for catching that....sorry about dropping out that divsion sign!!

Mea culpa!!