Question

What is the geometric mean of the sequence 1, 3, 9, 27, 81,... 3ⁿ?

• A. \(3^{\frac{n(n+1)}{2}}\)
• B. \(3^{\frac{n}{2}}\)
• C. \(3^{n}\)
• D. \(3^{2n}\)
• E. \(3^{n(n+1)}\)

Answer 1

The Correct Option is B

To find the geometric mean of a sequence, we use the formula for the geometric mean of \( n \) numbers. For a sequence of \( n \) terms \( a_1, a_2, \ldots, a_n \), the geometric mean \( G \) is given by: 

\(G = \sqrt[n]{a_1 \cdot a_2 \cdot \ldots \cdot a_n}\)

In the given sequence \( 1, 3, 9, 27, 81, \ldots, 3^n \), each term can be represented as \( 3^k \), where \( k \) ranges from 0 to \( n \).

Thus, the terms of the sequence are: \( 3^0, 3^1, 3^2, \ldots, 3^n \).

The product of these terms is:

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

Using the property of exponents, we can simplify the product:

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

The sum of the first \( n+1 \) natural numbers (from 0 to \( n \)) is given by the formula:

\(\text{Sum} = \frac{n(n+1)}{2}\)

Therefore, the product becomes:

\(P = 3^{\frac{n(n+1)}{2}}\)

Now, we find the geometric mean:

\(G = \sqrt[n+1]{3^{\frac{n(n+1)}{2}}}\)

\(G = 3^{\frac{\frac{n(n+1)}{2}}{n+1}}\)

Simplifying the exponent:

\(\frac{n(n+1)}{2(n+1)} = \frac{n}{2}\)

Therefore, the geometric mean is:

\(G = 3^{\frac{n}{2}}\)

Hence, the correct answer is \(3^{\frac{n}{2}}\).