Understanding the Geometric Mean Problem

The question asks us to find the value of a specific expression involving the geometric mean (G) of a sequence of numbers: 1, 2, 2², 2³, ..., 2^n-1.

First, let's understand what a geometric mean is and identify the numbers in the sequence.

What is a Geometric Mean?

The geometric mean of a set of 'n' positive numbers is the n-th root of their product.

For numbers \(a_1, a_2, \ldots, a_n\), the geometric mean G is given by:

$$G = \sqrt[n]{a_1 \cdot a_2 \cdot \cdots \cdot a_n}$$

Identifying the Sequence and Calculating Product

The given numbers are 1, 2, 2², 2³, ..., 2^n-1. These numbers can be written in the form of powers of 2:

• 1 = 2⁰
• 2 = 2¹
• 2²
• ...
• 2^n-1

There are exactly 'n' numbers in this sequence (from the power 0 up to n-1). To find the geometric mean G, we first need to calculate the product of these 'n' numbers:

Product \(P = 1 \cdot 2 \cdot 2^2 \cdot 2^3 \cdot \ldots \cdot 2^{n-1}\)

Writing this using powers of 2:

\(P = 2^0 \cdot 2^1 \cdot 2^2 \cdot 2^3 \cdot \ldots \cdot 2^{n-1}\)

When multiplying powers with the same base, we add the exponents:

\(P = 2^{(0 + 1 + 2 + 3 + \ldots + (n-1))}\)

The sum of the exponents is the sum of the first \(n\) non-negative integers, which is equivalent to the sum of the first \(n-1\) positive integers:

Sum of exponents \(= 0 + 1 + 2 + \ldots + (n-1)\)

Using the formula for the sum of the first \(k\) integers, \(1 + 2 + \ldots + k = \frac{k(k+1)}{2}\), here \(k = n-1\):

Sum of exponents \(= \frac{(n-1)((n-1)+1)}{2} = \frac{(n-1)n}{2}\)

So, the product \(P = 2^{\frac{n(n-1)}{2}}\).

Calculating the Geometric Mean G

The geometric mean G is the n-th root of the product P:

$$G = \sqrt[n]{P} = \sqrt[n]{2^{\frac{n(n-1)}{2}}}$$

This can be written using exponents as:

$$G = \left(2^{\frac{n(n-1)}{2}}\right)^{\frac{1}{n}}$$

Using the property \((a^m)^p = a^{mp}\):

$$G = 2^{\frac{n(n-1)}{2} \cdot \frac{1}{n}} = 2^{\frac{n-1}{2}}$$

So, the geometric mean \(G = 2^{\frac{n-1}{2}}\).

Evaluating the Expression 1 + 2log₂G

Now we substitute the value of G into the expression \(1 + 2\log_2G\):

Expression \(= 1 + 2\log_2\left(2^{\frac{n-1}{2}}\right)\)

Using the logarithm property \(\log_b(b^x) = x\), where \(b=2\) and \(x = \frac{n-1}{2}\):

$$\log_2\left(2^{\frac{n-1}{2}}\right) = \frac{n-1}{2}$$

Substituting this back into the expression:

Expression \(= 1 + 2 \cdot \left(\frac{n-1}{2}\right)\)

Now, simplify the expression:

Expression \(= 1 + (n-1)\)

Expression \(= 1 + n - 1\)

Expression \(= n\)

Therefore, the value of \(1 + 2\log_2G\) is \(n\).

Revision Table: Geometric Mean Calculation

Additional Information: Geometric Mean and Logarithms

Let's look at some related concepts that are useful when solving problems like this.

Properties of Logarithms

• \(\log_b(xy) = \log_b x + \log_b y\)
• \(\log_b(\frac{x}{y}) = \log_b x - \log_b y\)
• \(\log_b(x^p) = p \log_b x\)
• \(\log_b b = 1\)
• \(\log_b 1 = 0\)
• \(\log_b(b^x) = x\) (This was crucial in our solution)
• Change of base formula: \(\log_b a = \frac{\log_c a}{\log_c b}\)

Geometric Mean vs Arithmetic Mean

For a set of positive numbers, the geometric mean is always less than or equal to the arithmetic mean. Equality holds only if all the numbers in the set are identical.

• Arithmetic Mean (AM) of \(a_1, \ldots, a_n\) is \(\frac{a_1 + \dots + a_n}{n}\).
• Geometric Mean (GM) of \(a_1, \ldots, a_n\) is \(\sqrt[n]{a_1 \cdots a_n}\).
• AM \(\ge\) GM

Arithmetic Series Sum

The sum of an arithmetic series with 'k' terms, first term 'a', and last term 'l' is given by \(S_k = \frac{k}{2}(a+l)\). For the sum of the first \(n-1\) positive integers (1 to n-1), \(k=n-1\), \(a=1\), \(l=n-1\). The sum is \(\frac{n-1}{2}(1 + n-1) = \frac{n-1}{2}(n)\), which matches our result \(\frac{n(n-1)}{2}\). Adding 0 at the beginning doesn't change the sum.

By carefully calculating the geometric mean using properties of exponents and then using the properties of logarithms, we found that the expression \(1 + 2\log_2G\) simplifies to \(n\).