Problem 81 - Algebra and Trigonometry

Question

Determine whether the given sequence is arithmetic, geometric, or neither. If the sequence is arithmetic, find the common difference; if it is geometric, find the common ratio. If the sequence is arithmetic or geometric, find the sum of the first 50 terms. $$ \left\\{3^{n / 2}\right\\} $$

Step-by-Step Solution

Answer

Geometric with common ratio \( 3^{1/2} \); Sum of first 50 terms is \( 3^{1/2} \frac{3^{25} - 1}{3^{1/2} - 1} \).

Step 1: Identify the Sequence Type

The given sequence is \( \{3^{n / 2}\} \). To determine the type of the sequence, analyze the ratio of consecutive terms. Let’s compute the first few terms by substituting values of \( n \):\( n = 1 \Rightarrow a_1 = 3^{1 / 2} \)\( n = 2 \Rightarrow a_2 = 3^{2 / 2} = 3 \)\( n = 3 \Rightarrow a_3 = 3^{3 / 2} \)\( n = 4 \Rightarrow a_4 = 3^2 = 9 \).

Step 2: Check for Common Ratio

Calculate the common ratio \( r \) by dividing each term by its preceding term: \( r = \frac{a_{n+1}}{a_n} \). \( r_1 = \frac{a_2}{a_1} = \frac{3}{3^{1 / 2}} = 3^{1/2} \) \( r_2 = \frac{a_3}{a_2} = \frac{3^{3/2}}{3} = 3^{1/2} \) \( r_3 = \frac{a_4}{a_3} = \frac{9}{3^{3/2}} = 3^{1/2} \). Since the common ratio \( r = 3^{1/2} \) remains consistent, it confirms that the given sequence is geometric.

Step 3: Calculate the Sum of First 50 Terms

For a geometric sequence with first term \( a = 3^{1/2} \) and common ratio \( r = 3^{1/2} \), the sum of the first \( n \) terms is given by: \[ S_n = a \frac{r^n - 1}{r - 1} \]. Substituting \( a = 3^{1/2}, r = 3^{1/2}, n = 50 \): \[ S_{50} = 3^{1/2} \frac{(3^{1/2})^{50} - 1}{3^{1/2} - 1} \]. Simplify the terms: \[ (3^{1/2})^{50} = 3^{25} \] thus \[ S_{50} = 3^{1/2} \frac{3^{25} - 1}{3^{1/2} - 1} \].

Key Concepts

Common Ratio

In a geometric sequence, the common ratio is a constant value that you multiply each term by in order to get the next term. To identify the common ratio, you divide any term in the sequence by the preceding term. For instance, if the sequence is given as \( \{3^{n/2}\} \), we can compute the first few terms to determine the type of sequence.

Consider:

• When \( n = 1 \), \( a_1 = 3^{1/2} \)
• When \( n = 2 \), \( a_2 = 3 \)
• When \( n = 3 \), \( a_3 = 3^{3/2} \)
• When \( n = 4 \), \( a_4 = 3^2 = 9 \)

Now, calculate the common ratio \( r \) by dividing each term by its preceding term:

• \( r_1 = \frac{a_2}{a_1} = \frac{3}{3^{1/2}} = 3^{1/2} \)
• \( r_2 = \frac{a_3}{a_2} = \frac{3^{3/2}}{3} = 3^{1/2} \)
• \( r_3 = \frac{a_4}{a_3} = \frac{9}{3^{3/2}} = 3^{1/2} \)

Since the common ratio \( r = 3^{1/2} \) remains consistent, the given sequence is geometric.

Sum of Geometric Series

The sum of a geometric series is found using the formula: \[ S_n = a \frac{r^n - 1}{r - 1} \] Here, \( S_n \) is the sum of the first \( n \) terms, \( a \) is the first term, and \( r \) is the common ratio. From our sequence \( \{3^{n / 2}\} \):

• First term \( a = 3^{1/2} \)
• Common ratio \( r = 3^{1/2} \)
• \( n = 50 \)

Now, substitute these values into the formula: \[ S_{50} = 3^{1/2} \frac{(3^{1/2})^{50} - 1}{3^{1/2} - 1} \]
Simplify the expression inside the exponent: \[ (3^{1/2})^{50} = 3^{25} \] thus, \[ S_{50} = 3^{1/2} \frac{3^{25} - 1}{3^{1/2} - 1} \] Plug in these values to calculate the exact sum of the first 50 terms.

Geometric Sequence

A geometric sequence is a sequence of numbers where each term after the first is found by multiplying the previous term by a constant value, called the common ratio. The general form of a geometric sequence can be written as:

• First term \( a \)
• Second term \( ar \)
• Third term \( ar^2 \)
• And so on... \( ar^{n-1} \)

For instance, the sequence \( \{3^{n/2}\} \) can be evaluated:

For \( n = 1 \), \( a_1 = 3^{1/2} \)
For \( n = 2 \), \( a_2 = 3 \)
For \( n = 3 \), \( a_3 = 3^{3/2} \)
For \( n = 4 \), \( a_4 = 9 \)

We see that each term is connected by the common ratio \( 3^{1/2} \), meaning the given sequence is geometric.