What is the sum of the geometric series

2343 / 256

Explanation

I will do this for you in two forms: 1) adding each term, and 2) using the general formula for the sum of geometric series.

1) Adding the terms:

 4
∑ 3 (3/4)^i = 3 (3/4)^0 + 3 (3/4)^1 + 3 (3/4)^2 + 3 (3/4)^3 + 3 (3/4)^4
i=0

= 3 + 9/4 + 27/16 + 81/64 + 243/256 = [256*3 + 27*16 + 64*9 + 4*81 + 243] / 256 =

= 2343 / 256

2) Using the formula:

n-1
∑ A (r^i) = A [1 - r^(n) ] / [ 1 - r]
i=0

Here n - 1 = 4 => n = 5

r = 3/4

A = 3

Therefore the sum is 3 [ 1 - (3/4)^5 ] / [ 1 - (3/4) ] =

= 3 [ 1 - (3^5) / (4^5) ] / [ 1/4 ] = 3 { [ (4^5) - (3^5) ] / (4^5) } / {1/4} =

= (3 * 781) / (4^5) / (1/4) =  3 * 781 / (4^4) = 2343 / 256

So, no doubt, the answer is 2343 / 256