Understanding the Problem: Harmonic Mean of a Geometric Sequence

The question asks us to find the value of the expression $\frac{n}{H}$, where $H$ is the harmonic mean of a specific sequence of $n$ numbers. The sequence is given as $1, 2, 2^2, 2^3, \dots, 2^{n-1}$. Let's identify these numbers and the concept of the harmonic mean.

The sequence is $x_1=1=2^0$, $x_2=2=2^1$, $x_3=2^2$, and so on, up to $x_n=2^{n-1}$. This is a geometric progression with the first term $a=1$ and the common ratio $r=2$. There are exactly $n$ terms in this sequence.

The harmonic mean ($H$) of $n$ numbers $x_1, x_2, \dots, x_n$ is defined by the formula:

\begin{equation*} H = \frac{n}{\frac{1}{x_1} + \frac{1}{x_2} + \dots + \frac{1}{x_n}} = \frac{n}{\sum_{i=1}^{n} \frac{1}{x_i}} \end{equation*}

Calculating the Sum of Reciprocals

To find the harmonic mean $H$, we first need to calculate the sum of the reciprocals of the numbers in the sequence:

\begin{equation*} \sum_{i=1}^{n} \frac{1}{x_i} = \frac{1}{1} + \frac{1}{2} + \frac{1}{2^2} + \dots + \frac{1}{2^{n-1}} \end{equation*}

Let's write out the first few terms to see the pattern:

• Term 1: $\frac{1}{1} = \left(\frac{1}{2}\right)^0$
• Term 2: $\frac{1}{2} = \left(\frac{1}{2}\right)^1$
• Term 3: $\frac{1}{2^2} = \left(\frac{1}{2}\right)^2$
• ...
• Term n: $\frac{1}{2^{n-1}} = \left(\frac{1}{2}\right)^{n-1}$

The sum of reciprocals is $\sum_{i=1}^{n} \left(\frac{1}{2}\right)^{i-1}$. This is a geometric series with:

• First term $A = \left(\frac{1}{2}\right)^{1-1} = \left(\frac{1}{2}\right)^0 = 1$
• Common ratio $R = \frac{1}{2}$
• Number of terms $N = n$

The sum of a finite geometric series is given by $S_N = A \frac{1-R^N}{1-R}$. Applying this formula:

\begin{align*} \sum_{i=1}^{n} \frac{1}{2^{i-1}} &= 1 \cdot \frac{1 - \left(\frac{1}{2}\right)^n}{1 - \frac{1}{2}} \\ &= \frac{1 - \frac{1}{2^n}}{\frac{1}{2}} \\ &= 2 \left(1 - \frac{1}{2^n}\right) \\ &= 2 \left(\frac{2^n - 1}{2^n}\right) \\ &= \frac{2(2^n - 1)}{2^n} \\ &= \frac{2^n - 1}{2^{n-1}}\end{align*}

So, the sum of the reciprocals is $\frac{2^n - 1}{2^{n-1}}$.

Calculating the Harmonic Mean (H)

Now we can substitute this sum back into the formula for the harmonic mean $H$:

\begin{equation*} H = \frac{n}{\sum_{i=1}^{n} \frac{1}{x_i}} = \frac{n}{\frac{2^n - 1}{2^{n-1}}} \end{equation*}

To simplify, we multiply $n$ by the reciprocal of the denominator:

\begin{equation*} H = n \cdot \frac{2^{n-1}}{2^n - 1} = \frac{n \cdot 2^{n-1}}{2^n - 1} \end{equation*}

Finding the Value of n/H

The question asks for the value of $\frac{n}{H}$. We have the expression for $H$, so we can substitute it:

\begin{equation*} \frac{n}{H} = \frac{n}{\frac{n \cdot 2^{n-1}}{2^n - 1}} \end{equation*}

Again, we multiply $n$ by the reciprocal of the denominator:

\begin{equation*} \frac{n}{H} = n \cdot \frac{2^n - 1}{n \cdot 2^{n-1}} \end{equation*}

The $n$ terms in the numerator and denominator cancel out:

\begin{equation*} \frac{n}{H} = \frac{2^n - 1}{2^{n-1}} \end{equation*}

We can simplify this expression further by separating the terms in the numerator:

\begin{align*} \frac{2^n - 1}{2^{n-1}} &= \frac{2^n}{2^{n-1}} - \frac{1}{2^{n-1}} \\ &= 2^{n - (n-1)} - \frac{1}{2^{n-1}} \\ &= 2^1 - \frac{1}{2^{n-1}} \\ &= 2 - \frac{1}{2^{n-1}}\end{align*}

So, the value of $\frac{n}{H}$ is $2 - \frac{1}{2^{n-1}}$.

Comparing with Options

Let's compare our result with the given options:

Our calculated value $\(2-\frac{1}{2^{n-1}}\)$ matches Option 2.

Revision Table: Harmonic Mean Calculation

Additional Information: Means and Series

This problem combines concepts from means (specifically the harmonic mean) and sequences/series (specifically the geometric series).

• Harmonic Mean (HM): The HM is one of the three classical Pythagorean means (Arithmetic Mean, Geometric Mean, and Harmonic Mean). It is often used when dealing with rates or ratios. For two numbers $a$ and $b$, the HM is $\frac{2}{\frac{1}{a} + \frac{1}{b}} = \frac{2ab}{a+b}$. For $n$ numbers, it's $\frac{n}{\sum \frac{1}{x_i}}$. The harmonic mean is always less than or equal to the geometric mean, which is less than or equal to the arithmetic mean (for positive numbers).
• Geometric Sequence: A sequence where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. The form is $a, ar, ar^2, ar^3, \dots$.
• Geometric Series: The sum of the terms of a geometric sequence. The sum of the first $n$ terms of a geometric series with first term $a$ and common ratio $r$ is $S_n = a \frac{1-r^n}{1-r}$ (when $r \neq 1$). In our sum of reciprocals, the first term was 1 and the common ratio was 1/2.
• Understanding how to sum geometric series is crucial for calculating harmonic means of sequences like the one presented in the problem.