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).