Let `a_(0)=0` and `a_(n)=3a_(n-1)+1` for `n ge 1`. Then the remainder obtained dividing `a_(2010)` by `11` is

Options

A. `0`
B. `7`
C. `3`
D. `4`

Explanation

To solve the problem, we will first derive a general formula for the sequence defined by \( a_n = 3a_{n-1} + 1 \) with the initial condition \( a_0 = 0 \). Then, we will find \( a_{2010} \) and compute the remainder when it is divided by 11.

### Step 1: Calculate the first few terms of the sequence

1. **Initial term**:
\[ a_0 = 0 \]

2. **First term**:
\[ a_1 = 3a_0 + 1 = 3 \cdot 0 + 1 = 1 \]

3. **Second term**:
\[ a_2 = 3a_1 + 1 = 3 \cdot 1 + 1 = 4 \]

4. **Third term**:
\[ a_3 = 3a_2 + 1 = 3 \cdot 4 + 1 = 12 + 1 = 13 \]

5. **Fourth term**:
\[ a_4 = 3a_3 + 1 = 3 \cdot 13 + 1 = 39 + 1 = 40 \]

6. **Fifth term**:
\[ a_5 = 3a_4 + 1 = 3 \cdot 40 + 1 = 120 + 1 = 121 \]

### Step 2: Identify a pattern or formula

From the calculations, we have:
- \( a_0 = 0 \)
- \( a_1 = 1 \)
- \( a_2 = 4 \)
- \( a_3 = 13 \)
- \( a_4 = 40 \)
- \( a_5 = 121 \)

### Step 3: Find a general formula for \( a_n \)

We can observe that the recurrence relation \( a_n = 3a_{n-1} + 1 \) can be solved using the method of iteration or by recognizing it as a geometric series.

We can express \( a_n \) in terms of a closed formula:
\[ a_n = 3^n - 1 \]

### Step 4: Calculate \( a_{2010} \)

Using the formula:
\[ a_{2010} = 3^{2010} - 1 \]

### Step 5: Find \( a_{2010} \mod 11 \)

To find \( a_{2010} \mod 11 \), we first need to calculate \( 3^{2010} \mod 11 \).

Using Fermat's Little Theorem, since 11 is prime:
\[ 3^{10} \equiv 1 \mod 11 \]

Now, we find \( 2010 \mod 10 \):
\[ 2010 \mod 10 = 0 \]

Thus:
\[ 3^{2010} \equiv (3^{10})^{201} \equiv 1^{201} \equiv 1 \mod 11 \]

### Step 6: Calculate \( a_{2010} \mod 11 \)

Now substituting back:
\[ a_{2010} \equiv 3^{2010} - 1 \equiv 1 - 1 \equiv 0 \mod 11 \]

### Final Answer

The remainder obtained when dividing \( a_{2010} \) by 11 is:
\[ \boxed{0} \]