Find the sum to n terms of the sequence 3,33,333,3333...

To find the sum of the first *n* terms of the sequence 3, 33, 333, 3333, ..., we first need to identify the general form of each term.

### Step 1: General form of the terms
The first few terms of the sequence are:
- 3
- 33
- 333
- 3333
- ...

Each term in the sequence consists of repeated digits "3." We can express these terms mathematically as follows:

- 3 = 3
- 33 = 30 + 3 = 3(10 + 1)
- 333 = 300 + 30 + 3 = 3(100 + 10 + 1)
- 3333 = 3000 + 300 + 30 + 3 = 3(1000 + 100 + 10 + 1)

In general, the *k*-th term of the sequence can be written as:
- \( T_k = 3 \times (10^{k-1} + 10^{k-2} + \cdots + 10^0) \)
This is a sum of powers of 10, multiplied by 3.

### Step 2: Sum of the series
We are asked to find the sum of the first *n* terms, i.e., \( S_n = T_1 + T_2 + T_3 + \cdots + T_n \).

The *k*-th term, \( T_k \), is:
- \( T_k = 3 \times (10^{k-1} + 10^{k-2} + \cdots + 10^0) \)

We can express this as:
- \( T_k = 3 \times \frac{10^k - 1}{9} \) (this is the sum of a geometric series)

Thus, the sum of the first *n* terms, \( S_n \), is:
- \( S_n = 3 \times \sum_{k=1}^{n} \frac{10^k - 1}{9} \)

### Step 3: Simplifying the sum
We can split the sum into two parts:
- \( S_n = \frac{3}{9} \times \sum_{k=1}^{n} (10^k - 1) \)
- \( S_n = \frac{1}{3} \times \left( \sum_{k=1}^{n} 10^k - \sum_{k=1}^{n} 1 \right) \)

The sum of powers of 10 is a geometric series:
- \( \sum_{k=1}^{n} 10^k = 10 + 10^2 + 10^3 + \cdots + 10^n = \frac{10(10^n - 1)}{9} \)

The second sum is simply:
- \( \sum_{k=1}^{n} 1 = n \)

Thus, the sum \( S_n \) becomes:
- \( S_n = \frac{1}{3} \left( \frac{10(10^n - 1)}{9} - n \right) \)

### Final Answer:
The sum of the first *n* terms of the sequence is:
- \( S_n = \frac{1}{3} \left( \frac{10(10^n - 1)}{9} - n \right) \)