Evaluate Σ from n=1 to infinity of 3/2^n-1
a) 12
b) 6
c) 3
d) 0 ----> my answer. Can you check for me, please? Thanks in advance
you just have a geometric series where
a = 3
r = 1/2
S = 3/(1 - 1/2) = 6
The nth term converges to zero, but the sum does not
To evaluate the series Σ from n=1 to infinity of 3/2^n-1, we can first simplify the expression 3/2^n-1.
Recall that for any positive integer n, 2^n-1 equals 2^(n-1). So, we can rewrite the expression as 3/2^(n-1).
The series becomes: Σ from n=1 to infinity of 3/2^(n-1).
Now, let's analyze the terms of the series:
When n = 1, we have 3/2^(1-1) = 3/2^0 = 3/1 = 3.
When n = 2, we have 3/2^(2-1) = 3/2^1 = 3/2 = 1.5.
When n = 3, we have 3/2^(3-1) = 3/2^2 = 3/4 = 0.75.
And so on.
We can observe that every term of the series is half of the previous term, which means it is a geometric series with a common ratio of 1/2.
To calculate the sum of an infinite geometric series, we can use the formula S = a / (1 - r), where a is the first term and r is the common ratio.
In this case, a = 3 (the first term) and r = 1/2 (the common ratio).
Applying the formula, we have S = 3 / (1 - 1/2) = 3 / (1/2) = 3 * (2/1) = 6.
Therefore, the sum of the series Σ from n=1 to infinity of 3/2^(n-1) is 6.
Hence, the correct answer is b) 6.