Question

Evaluate the summation from n equals 1 to infinity of the quotient of 3 and 2 raised the quantity n minus 1 power

Answer 1

Answer:

3*∑ (1/2)^n = 6

Step-by-step 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