1. Interpret the expression:

The sum is $1+3+3^2+3^3+\cdots+3^{2021}.$

We need the remainder when this sum is divided by $50$.

2. Write it as a geometric series:

$S=1+3+3^2+\cdots+3^{2021}.$

Using the geometric series formula, $S=\frac{3^{2022}-1}{3-1}=\frac{3^{2022}-1}{2}.$

But working directly modulo $50$ with division by $2$ is inconvenient, so instead we find the sum modulo $50$ using powers of $3$ modulo $50$.

3. Find the pattern of $3^n \pmod{50}$:

Compute a few powers: $3^1=3,$ $3^2=9,$ $3^3=27,$ $3^4=81\equiv 31 \pmod{50},$ $3^5=93\equiv 43 \pmod{50},$ $3^6=129\equiv 29 \pmod{50},$ $3^7=87\equiv 37 \pmod{50},$ $3^8=111\equiv 11 \pmod{50},$ $3^9=33 \pmod{50},$ $3^{10}=99\equiv 49 \pmod{50},$ $3^{11}=147\equiv 47 \pmod{50},$ $3^{12}=141\equiv 41 \pmod{50},$ $3^{13}=123\equiv 23 \pmod{50},$ $3^{14}=69\equiv 19 \pmod{50},$ $3^{15}=57\equiv 7 \pmod{50},$ $3^{16}=21 \pmod{50},$ $3^{17}=63\equiv 13 \pmod{50},$ $3^{18}=39 \pmod{50},$ $3^{19}=117\equiv 17 \pmod{50},$ $3^{20}=51\equiv 1 \pmod{50}.$

So the powers repeat with period $20$.

4. Sum one full cycle:

We need $1+3+3^2+\cdots+3^{19} \pmod{50}.$ Since $1+3+3^2+\cdots+3^{19}=\frac{3^{20}-1}{3-1}=\frac{3^{20}-1}{2}.$ Now, $3^{20}\equiv 1 \pmod{50},$ but better compute modulo $100$ so division by $2$ is valid:

Using powers modulo $100$, $3^{20}\equiv 1 \pmod{100}$ (since $3^{20}=3486784401$, ending in $01$). Thus, $3^{20}-1\equiv 0 \pmod{100},$ so $\frac{3^{20}-1}{2}\equiv 0 \pmod{50}.$

Hence, $1+3+3^2+\cdots+3^{19}\equiv 0 \pmod{50}.$

5. Break the full sum into complete cycles:

There are terms from exponent $0$ to $2021$, i.e. $2022$ terms.

Since the period is $20$, $2022=20\cdot 101+2.$

So, $S=(1+3+3^2+\cdots+3^{19})+\cdots+\text{(101 such full blocks)}+3^{2020}+3^{2021}.$

Each full block contributes remainder $0$ modulo $50$.

Thus, $S\equiv 3^{2020}+3^{2021} \pmod{50}.$

Since $2020\equiv 0 \pmod{20}$ and $2021\equiv 1 \pmod{20}$, $3^{2020}\equiv 1, \qquad 3^{2021}\equiv 3 \pmod{50}.$

Therefore, $S\equiv 1+3=4 \pmod{50}.$

6. Final answer:

The remainder is $\boxed{4}.$