`1+3+3^2+ . . . +3^(2021)` is divided by 50 then what will be remainder ?

Answer

To solve the problem of finding the remainder when the sum \( S = 1 + 3 + 3^2 + \ldots + 3^{2021} \) is divided by 50, we can follow these steps:

### Step 1: Identify the series
The given series is a geometric progression (GP) where:
- First term \( a = 1 \)
- Common ratio \( r = 3 \)
- Number of terms \( n = 2022 \) (from \( 3^0 \) to \( 3^{2021} \))

### Step 2: Use the formula for the sum of a GP
The sum of the first \( n \) terms of a geometric series can be calculated using the formula:
\[ S_n = a \frac{r^n - 1}{r - 1} \]
Substituting the values we have:
\[ S = 1 \cdot \frac{3^{2022} - 1}{3 - 1} = \frac{3^{2022} - 1}{2} \]

### Step 3: Find \( S \mod 50 \)
To find the remainder of \( S \) when divided by 50, we need to compute \( \frac{3^{2022} - 1}{2} \mod 50 \).

### Step 4: Calculate \( 3^{2022} \mod 50 \)
To simplify \( 3^{2022} \mod 50 \), we can use Euler's theorem. First, we find \( \phi(50) \):
\[ \phi(50) = 50 \left(1 - \frac{1}{2}\right)\left(1 - \frac{1}{5}\right) = 50 \cdot \frac{1}{2} \cdot \frac{4}{5} = 20 \]
According to Euler's theorem, since \( \gcd(3, 50) = 1 \):
\[ 3^{20} \equiv 1 \mod 50 \]
Now, we need to find \( 2022 \mod 20 \):
\[ 2022 \div 20 = 101 \quad \text{(remainder 2)} \]
Thus, \( 2022 \equiv 2 \mod 20 \).

Therefore:
\[ 3^{2022} \equiv 3^2 \mod 50 \]
Calculating \( 3^2 \):
\[ 3^2 = 9 \]

### Step 5: Substitute back into the sum
Now we substitute back into our sum:
\[ S = \frac{3^{2022} - 1}{2} \equiv \frac{9 - 1}{2} \mod 50 \]
This simplifies to:
\[ S \equiv \frac{8}{2} \mod 50 \equiv 4 \mod 50 \]

### Conclusion
The remainder when \( 1 + 3 + 3^2 + \ldots + 3^{2021} \) is divided by 50 is:
\[ \boxed{4} \]