How do you find the sum of the geometric series #1/9-1/3+1-...# to 6 terms?
1 Answer
Jan 16, 2017
Explanation:
The sum to n terms of a geometric progression is.
#color(red)(bar(ul(|color(white)(2/2)color(black)(S_n=(a(1-r^n))/(1-r)=(a(r^n-1))/(r-1))color(white)(2/2)|)))#
where a is the first term and r, the common ratio.
#"here " a=1/9" and " r=(a_2)/(a_1)=(1/9)/(-1/3)=-3#
#"thus for " n=6#
#S_6=(1/9((-3)^6-1))/(-4)#
#=(1/9xx728)/-4=-182/9#
