Question

For all \( n \in \mathbb{N} \), which of the following is less than or equal to \( \frac{3^n - 1}{2} \)?

• A. \( n^2 \left(2^{n/2} \right) \)
• B. \( n^2 \left(\frac{n - 1}{3^2} \right) \)
• C. \( n^3 \left(\frac{n - 1}{3^2} \right) \)
• D. \( n \left(\frac{n - 1}{3^2} \right) \)

Hint:

When comparing polynomial expressions with exponential ones, remember that exponentials grow significantly faster as \( n \) increases. Use test values to verify inequalities.

Answer 1

The Correct Option is D

We are asked to find a function \( f(n) \) such that for all \( n \in \mathbb{N} \), \[ f(n) \leq \frac{3^n - 1}{2} \] Step 1: Understand the Growth of \( \frac{3^n - 1}{2} \) 
This is an exponential function with base 3. As \( n \) increases, \( \frac{3^n - 1}{2} \) increases very rapidly. 
Step 2: Examine the Options 
Let’s denote \( f(n) = n \cdot \frac{n - 1}{3^2} = \frac{n(n - 1)}{9} \) This is a quadratic function in \( n \), so its growth is much slower than that of the exponential function \( \frac{3^n - 1}{2} \)
Step 3: Check a few values to verify: For \( n = 1 \): \[ \frac{3^1 - 1}{2} = 1, f(1) = \frac{1(0)}{9} = 0 \] For \( n = 2 \): \[ \frac{9 - 1}{2} = 4, f(2) = \frac{2(1)}{9} = \frac{2}{9} \] For \( n = 3 \): \[ \frac{27 - 1}{2} = 13, f(3) = \frac{3(2)}{9} = \frac{6}{9} = \frac{2}{3} \] In all cases, \( f(n)<\frac{3^n - 1}{2} \) 
Conclusion: Among all options, only Option (D) satisfies the inequality for all natural numbers.