If `a_(n+1)=a_(n-1)+2a_(n)` for `n=2,3,4, . . .` and `a_(1)=1` and `a_(2)=1`, then `a_(5)`=

Options

A. 7
B. 11
C. 17
D. 21

Explanation

To solve the recurrence relation given by \( a_{n+1} = a_{n-1} + 2a_n \) for \( n = 2, 3, 4, \ldots \) with initial conditions \( a_1 = 1 \) and \( a_2 = 1 \), we will calculate the values of \( a_3 \), \( a_4 \), and \( a_5 \) step by step.

### Step 1: Calculate \( a_3 \)
Using the recurrence relation for \( n = 2 \):
\[ a_{3} = a_{1} + 2a_{2} \]
Substituting the known values:
\[ a_{3} = 1 + 2 \times 1 = 1 + 2 = 3 \]

### Step 2: Calculate \( a_4 \)
Now, using the recurrence relation for \( n = 3 \):
\[ a_{4} = a_{2} + 2a_{3} \]
Substituting the known values:
\[ a_{4} = 1 + 2 \times 3 = 1 + 6 = 7 \]

### Step 3: Calculate \( a_5 \)
Next, using the recurrence relation for \( n = 4 \):
\[ a_{5} = a_{3} + 2a_{4} \]
Substituting the known values:
\[ a_{5} = 3 + 2 \times 7 = 3 + 14 = 17 \]

Thus, the value of \( a_5 \) is \( \boxed{17} \).