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!!