Question

How do you find the sum of the geometric series `Sigma 3* 2^(n-1)` from n=1 to 20?

Answer 1

`3145725`

Explanation:

`sum_(k=1)^n ar^(k-1)= a(r^n -1)/(r -1 )`

`sum_(k=1) ^n 3. 2^(n-1) = 3((2)^20 -1)/(2 -1)` where `a = 3, r` = 2, n = 20#

`sum_(k=1) ^20 = 3((2)^20 -1)/(2 -1)`

`= 3((2)^20 -1)/(2 -1)`

`= 3(1048576 -1)/1`

`= 3(1048575) = 3145725`