The remainder on dividing `1+3+3^(2)+3^(3) +……..+ 3^(2021) " by " 50 ` is ______.

Answer

To find the remainder when dividing the sum \( S = 1 + 3 + 3^2 + 3^3 + \ldots + 3^{2021} \) by 50, we can recognize that this is a geometric series.

### Step-by-Step Solution:

1. **Identify the series**:
The series can be expressed as:
\[ S = 3^0 + 3^1 + 3^2 + \ldots + 3^{2021} \]
This is a geometric series where the first term \( a = 1 \) (or \( 3^0 \)), the common ratio \( r = 3 \), and the number of terms \( n = 2022 \) (from \( 0 \) to \( 2021 \)).

2. **Use the formula for the sum of a geometric series**:
The sum of the first \( n \) terms of a geometric series is given by:
\[ S_n = a \frac{r^n - 1}{r - 1} \]
Substituting the values we have:
\[ S = 1 \cdot \frac{3^{2022} - 1}{3 - 1} = \frac{3^{2022} - 1}{2} \]

3. **Calculate \( 3^{2022} \mod 50 \)**:
To find the remainder of \( S \) when divided by 50, we first need to compute \( 3^{2022} \mod 50 \). We can use Euler's theorem, which states that if \( a \) and \( n \) are coprime, then:
\[ a^{\phi(n)} \equiv 1 \mod n \]
Here, \( n = 50 \) and \( \phi(50) = 50 \cdot (1 - \frac{1}{2}) \cdot (1 - \frac{1}{5}) = 20 \).

4. **Reduce the exponent modulo \( \phi(50) \)**:
Since \( 3 \) and \( 50 \) are coprime, we can reduce the exponent:
\[ 2022 \mod 20 = 2 \]
Therefore, \( 3^{2022} \equiv 3^2 \mod 50 \).

5. **Calculate \( 3^2 \mod 50 \)**:
\[ 3^2 = 9 \]

6. **Substitute back into the sum**:
Now substitute back into the sum:
\[ S \equiv \frac{3^{2022} - 1}{2} \mod 50 \]
Thus,
\[ S \equiv \frac{9 - 1}{2} \mod 50 \]
\[ S \equiv \frac{8}{2} \mod 50 \]
\[ S \equiv 4 \mod 50 \]

### Final Answer:
The remainder when \( 1 + 3 + 3^2 + \ldots + 3^{2021} \) is divided by 50 is \( \boxed{4} \).