Understanding the Arithmetico-Geometric Series Sum

The problem asks us to find the value of the summation \(\sum_{i=1}^n x_i f_i\), where the terms \(x_i\) and \(f_i\) are defined as follows:

• The sequence \(x_i\) is \(1, 2, 3, \dots, n\).
• The sequence \(f_i\) is \(1, 2^{-1}, 2^{-2}, \dots, 2^{-(n-1)}\).

Let's write out the terms of the sum:

The summation \(S\) can be expressed as:

This can be rewritten as:

Identifying the Series Type

This series is an example of an arithmetico-geometric series. This type of series is the product of terms from an arithmetic progression and a geometric progression.

• The arithmetic part consists of the terms: \(1, 2, 3, \dots, n\) (first term \(a=1\), common difference \(d=1\)).
• The geometric part consists of the terms: \(1, \frac{1}{2}, \frac{1}{2^2}, \dots, \frac{1}{2^{n-1}}\) (first term \(a'=1\), common ratio \(r = \frac{1}{2}\)).

Step-by-Step Calculation of the Sum

To find the sum \(S\), we can use a standard method for arithmetico-geometric series. Let the common ratio of the geometric part be \(r = \frac{1}{2}\).

1. Write the sum \(S\): \(S = 1 + \frac{2}{2} + \frac{3}{2^2} + \frac{4}{2^3} + \dots + \frac{n}{2^{n-1}} \quad (*)\)
2. Multiply the entire equation by the common ratio \(r = \frac{1}{2}\): \(\frac{1}{2}S = \frac{1}{2} + \frac{2}{2^2} + \frac{3}{2^3} + \frac{4}{2^4} + \dots + \frac{n}{2^n} \quad (**)\)
3. Subtract equation (**) from equation (*): \(S - \frac{1}{2}S = \left(1 + \frac{2}{2} + \frac{3}{2^2} + \dots + \frac{n}{2^{n-1}}\right) - \left(\frac{1}{2} + \frac{2}{2^2} + \dots + \frac{n-1}{2^{n-1}} + \frac{n}{2^n}\right)\) \(\frac{1}{2}S = 1 + \left(\frac{2}{2} - \frac{1}{2}\right) + \left(\frac{3}{2^2} - \frac{2}{2^2}\right) + \dots + \left(\frac{n}{2^{n-1}} - \frac{n-1}{2^{n-1}}\right) - \frac{n}{2^n}\) \(\frac{1}{2}S = 1 + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \dots + \frac{1}{2^{n-1}} - \frac{n}{2^n}\)
4. The terms \(1 + \frac{1}{2} + \frac{1}{2^2} + \dots + \frac{1}{2^{n-1}}\) form a finite geometric series with first term \(a'=1\), common ratio \(r=\frac{1}{2}\), and \(n\) terms. The sum of this geometric series is given by \(\frac{a'(1-r^n)}{1-r}\). Sum of geometric series = \(\frac{1 \left(1 - \left(\frac{1}{2}\right)^n\right)}{1 - \frac{1}{2}} = \frac{1 - \frac{1}{2^n}}{\frac{1}{2}} = 2 \left(1 - \frac{1}{2^n}\right) = 2 - \frac{2}{2^n} = 2 - \frac{1}{2^{n-1}}\).
5. Substitute this back into the equation for \(\frac{1}{2}S\): \(\frac{1}{2}S = \left(2 - \frac{1}{2^{n-1}}\right) - \frac{n}{2^n}\) \(\frac{1}{2}S = 2 - \frac{2}{2^n} - \frac{n}{2^n}\) \(\frac{1}{2}S = 2 - \frac{n+2}{2^n}\)
6. Now, solve for \(S\) by multiplying by 2: \(S = 2 \left(2 - \frac{n+2}{2^n}\right)\) \(S = 4 - \frac{2(n+2)}{2^n}\) \(S = 4 - \frac{n+2}{2^{n-1}}\)
7. To match the format of the options, express 4 with the denominator \(2^{n-1}\): \(4 = \frac{4 \cdot 2^{n-1}}{2^{n-1}} = \frac{2^2 \cdot 2^{n-1}}{2^{n-1}} = \frac{2^{n+1}}{2^{n-1}}\)
8. Substitute this back into the expression for \(S\): \(S = \frac{2^{n+1}}{2^{n-1}} - \frac{n+2}{2^{n-1}}\) \(S = \frac{2^{n+1} - (n+2)}{2^{n-1}}\) \(S = \frac{2^{n+1} - n - 2}{2^{n-1}}\)

Final Answer Verification

Comparing our derived sum \(S = \frac{2^{n+1} - n - 2}{2^{n-1}}\) with the given options, we find it matches option 2.