Question

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

Answer 1

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