Question

What is the sum of the first five terms of a geometric series with a1 = 6 and r = 1/3?

Answer 1

sum = a(1-r^n)/(1-r)
=6[1-(1/3)^5]/(1-1/3)
=6*(242/243)/(2/3)
= 242/27

Answer 2

or with the concept of geometry, u can times a1 by the r to get a2, and then times again a2 by r, you get a3 ( this is only for simple condition ). so
a2 = a1*r = 6*1/3 = 2;
a3 = a2*r = 2*1/3 = 2/3;
.
.
and i got a5 = 2/27;
then u can sum it, a1 + a2 + . . . + a5 = 242/27

But, if had some complex condition, u can use Callisto's solution, e.g such as sum first one hundred terms of geometric series ( u can use callisto's solution).

Thank you.