evaluate:
n sigma i=1 3/2^(i-1)
To evaluate the given sum, n Σ i=1 3/2^(i-1), we can first rewrite it using sigma notation:
n Σ i=1 3/2^(i-1) = 3/2^(1-1) + 3/2^(2-1) + 3/2^(3-1) + ... + 3/2^((n-1)-1)
Now, let's simplify each term:
3/2^(1-1) = 3/2^0 = 3/1 = 3
3/2^(2-1) = 3/2^1 = 3/2
3/2^(3-1) = 3/2^2
...
We can notice a pattern here. The general term of the sequence is given by:
3/2^(i-1)
Now, the sum can be rewritten as:
n Σ i=1 3/2^(i-1) = 3 + 3/2 + 3/2^2 + ... + 3/2^(n-1)
This is a geometric series with a common ratio of 1/2. The formula for the sum of a geometric series is:
S = a * (1 - r^n) / (1 - r)
where S is the sum of the series, a is the first term, r is the common ratio, and n is the number of terms.
In this case, a = 3, r = 1/2, and the number of terms is n. Plugging these values into the formula, we get:
S = 3 * (1 - (1/2)^n) / (1 - 1/2)
Simplifying further:
S = 3 * (1 - (1/2)^n) / (1/2)
Multiplying both the numerator and denominator by 2 to simplify:
S = 6 * (1 - (1/2)^n)
So, the evaluated sum of the expression n Σ i=1 3/2^(i-1) is 6 * (1 - (1/2)^n).