Problem 1290

1. **Identify the $n$-th term of the sequence:**
The given sequence is $1, 1+2, 1+2+2^2, \ldots, 1+2+\cdots+2^{k-1}, \ldots$. We need to find the $n$-th term of this sequence.

The $n$-th term can be written as:
$$a_n = \sum_{p=0}^{n-1} 2^p$$

2. **Simplify the $n$-th term:**
The sum of a geometric series $\sum_{p=0}^{n-1} 2^p$ is given by:
$$a_n = \sum_{p=0}^{n-1} 2^p = \frac{2^n - 1}{2 - 1} = 2^n - 1$$

3. **Find the sum of the first $n$ terms of the sequence:**
The sum of the first $n$ terms, $S_n$, is:
$$S_n = \sum_{k=1}^n a_k = \sum_{k=1}^n (2^k - 1)$$

4. Separate the sum into two parts:
$$S_n = \sum_{k=1}^n 2^k - \sum_{k=1}^n 1$$

5. Evaluate each part separately:
- The sum of the geometric series $\sum_{k=1}^n 2^k$ is:
$$\sum_{k=1}^n 2^k = 2 + 2^2 + 2^3 + \cdots + 2^n = 2(1 + 2 + 2^2 + \cdots + 2^{n-1}) = 2(2^n - 1) = 2^{n+1} - 2$$
- The sum of the constant series $\sum_{k=1}^n 1$ is:
$$\sum_{k=1}^n 1 = n$$

6. Combine the results:
$$S_n = (2^{n+1} - 2) - n = 2^{n+1} - 2 - n$$

7. Compare with the given form:
The given form is $2^{n+R} + Sn^2 + Tn + U$. By comparing:
$$2^{n+1} - 2 - n = 2^{n+1} + 0 \cdot n^2 - 1 \cdot n - 2$$

We can identify the constants:
$$R = 1, \quad S = 0, \quad T = -1, \quad U = -2$$

The final answer is $\boxed{ R = 1, S = 0, T = -1, U = -2 }$.