How do you find the sum of the geometric series #Sigma 4*3^(n-1)# from n=1 to 16?

1 Answer
Sep 10, 2017

16
#sum 4*3^(n-1) = 86093440#
n=1

Explanation:

16
#sum 4*3^(n-1) = ?#
n=1
1st term #a_1= 4*3^(1-1)=4 ; n=1# ,
2nd term #a_2= 4*3^(2-1)=12 ; n=2#
3rd term #a_3= 4*3^(3-1)=36 ; n=3#
4th term #a_4= 4*3^(4-1)=108 ; n=4#

So geometric series is # 4 ,12 , 36 ,108 ......# of which

1st term is #a_1=4# and common ratio is #r=3#

Sum formula of geometric series is # S_n= a_1 *( r^n-1)/(r-1)# or

# S_16 = a_1 * ( r^n-1)/(r-1) = 4* (3^16 -1)/(3-1)=86093440 #

16
#sum 4*3(n-1) = 86093440#[Ans]
n=1