Answer:
3*∑ (1/2)^n = 6
Explanation:
We have the summation from n = 1 to n = ∞, for:
∑ 3/(2)^(n - 1)
We know that the summation between k = 0, and k = N - 1 for:
∑ r^k = (1 - r^N)/(1 - r)
if we have the summation between k = 0 and k = ∞ - 1 = ∞
(here we used that ∞ is really big, then ∞ - 1 = ∞)
In the numerator we will have the term r^∞, if 0 < r < 1, then r^∞ = 0.
Then if we assume that 0 < r < 1 we can write:
∑ r^k = 1/(1 - r)
In our case, we can rewrite our summation as:
3*∑ (1/2)^n
for n = 0 to n = ∞
You can see that i changed the limits for n, this does not really matter because we are summing between a number and infinity, the only thing you need to take care is that now the power is n, instead of (n - 1)
Then we have r = (1/2) which is clearly smaller than 1.
then (1/2)
Then this summation is equal to:
3*∑ (1/2)^n = 3*( 1/(1 - 1/2)) = 3*( 1/( 2/2 - 1/2)) = 3*( 1/(1/2)) = 3*2 = 6
3*∑ (1/2)^n = 6