The number of `n` digit number formed by using digits `{1,2,3}` such that if `1` appears, it appears even number of times, is

Options

A. `2^(n)+1`
B. `(1)/(2)(3^(n)+1)`
C. `(1)/(2)(3^(n)-1)`
D. `(1)/(2)(2^(n)-1)`

Explanation

To solve the problem of finding the number of `n` digit numbers formed using the digits `{1, 2, 3}` such that if `1` appears, it appears an even number of times, we can break it down into several cases based on the occurrences of the digit `1`.

### Step-by-step Solution:

1. **Understanding the Cases**:
- We need to consider different cases based on how many times the digit `1` appears in the number. The digit `1` can appear `0, 2, 4, ..., n` times (if `n` is even) or `0, 2, 4, ..., n-1` times (if `n` is odd).

2. **Case 1: `1` appears 0 times**:
- If `1` appears 0 times, we can only use the digits `2` and `3`.
- The total number of `n` digit numbers formed using `2` and `3` is \(2^n\).

3. **Case 2: `1` appears 2 times**:
- If `1` appears 2 times, we need to choose 2 positions out of `n` for the digit `1`. This can be done in \(\binom{n}{2}\) ways.
- The remaining `n - 2` positions can be filled with either `2` or `3`, which gives us \(2^{n-2}\) combinations.
- Therefore, the total for this case is \(\binom{n}{2} \cdot 2^{n-2}\).

4. **Case 3: `1` appears 4 times**:
- If `1` appears 4 times, we choose 4 positions from `n`, which can be done in \(\binom{n}{4}\) ways.
- The remaining `n - 4` positions can again be filled with `2` or `3`, giving us \(2^{n-4}\) combinations.
- Thus, the total for this case is \(\binom{n}{4} \cdot 2^{n-4}\).

5. **Continuing the Pattern**:
- We can continue this pattern for all even occurrences of `1` up to `n`. The general formula for each case where `1` appears `2k` times (where \(k\) is a non-negative integer) is:
\[ \binom{n}{2k} \cdot 2^{n-2k} \]
- The maximum value of \(k\) is \(\lfloor n/2 \rfloor\).

6. **Summing Up All Cases**:
- The total number of valid `n` digit numbers is the sum of all the cases:
\[ S = \sum_{k=0}^{\lfloor n/2 \rfloor} \binom{n}{2k} \cdot 2^{n-2k} \]

7. **Using Binomial Theorem**:
- The sum \(S\) can be simplified using the binomial theorem. The expression can be related to the binomial expansion of \((1 + 2)^n\) and \((1 + 2)^n + (1 - 2)^n\). The even coefficients can be extracted from this expansion.
- The final result is:
\[ S = \frac{1}{2} \left(3^n + 1\right) \]

### Final Answer:
Thus, the number of `n` digit numbers formed by using digits `{1, 2, 3}` such that if `1` appears, it appears an even number of times is:
\[ \frac{3^n + 1}{2} \]