The first term of a sequence is 1, the second is 2 and every term is the sum of the two preceding terms.The `n^(th)` term is.

Options

A. `(1)/(2^(n+1) sqrt(5)) [( 1+ sqrt(5))^(n+1) + (1-sqrt(5))^(n+1)]`
B. `(1)/(2^(n+1) sqrt(5)) [( 1+ sqrt(5))^(n+1) - (1-sqrt(5))^(n+1)]`
C. `(1)/(2^(n+1) sqrt(5)) [( 1 - sqrt(5))^(n+1) + (1 - sqrt(5))^(n-1)]`
D. `(1)/(2^(n+1) sqrt(5)) [( 1+ sqrt(5))^(n-1) + (1-sqrt(5))^(n-1)]`

Answer

To find the nth term of the given sequence where the first term is 1, the second term is 2, and every subsequent term is the sum of the two preceding terms, we can follow these steps:

### Step 1: Identify the Sequence
The first two terms are given:
- \( a_1 = 1 \)
- \( a_2 = 2 \)

The subsequent terms are defined as:
- \( a_n = a_{n-1} + a_{n-2} \)

### Step 2: Calculate the First Few Terms
Let's calculate the first few terms of the sequence:
- \( a_3 = a_2 + a_1 = 2 + 1 = 3 \)
- \( a_4 = a_3 + a_2 = 3 + 2 = 5 \)
- \( a_5 = a_4 + a_3 = 5 + 3 = 8 \)
- \( a_6 = a_5 + a_4 = 8 + 5 = 13 \)

So, the first few terms are: 1, 2, 3, 5, 8, 13...

### Step 3: Recognize the Pattern
The sequence resembles the Fibonacci sequence, but it starts with different initial values. The Fibonacci sequence is defined as:
- \( F_1 = 1 \)
- \( F_2 = 1 \)
- \( F_n = F_{n-1} + F_{n-2} \)

However, in this case, the second term is 2 instead of 1.

### Step 4: Use the Fibonacci Formula
The nth term of the Fibonacci sequence can be expressed using Binet's formula:
\[ F_n = \frac{\phi^n - (1 - \phi)^n}{\sqrt{5}} \]
where \( \phi = \frac{1 + \sqrt{5}}{2} \) (the golden ratio).

### Step 5: Adjust for the Given Sequence
Since our sequence starts with 1 and 2, we can express the nth term as:
\[ a_n = F_{n} + F_{n-1} \]
where \( F_n \) is the nth Fibonacci number.

### Step 6: Write the nth Term
Thus, the nth term can be expressed as:
\[ a_n = \frac{\phi^n - (1 - \phi)^n}{\sqrt{5}} + \frac{\phi^{n-1} - (1 - \phi)^{n-1}}{\sqrt{5}} \]

### Step 7: Simplify
Combining the two terms gives us:
\[ a_n = \frac{\phi^n + \phi^{n-1} - (1 - \phi)^n - (1 - \phi)^{n-1}}{\sqrt{5}} \]

### Final Expression
This can be simplified further, but the main idea is that the nth term of the sequence can be expressed in terms of the Fibonacci numbers.