The sum of the first $n$ terms of the sequence $$1,1+2,1+2+2^2,\ldots,1+2+\cdots+2^{k-1},\ldots$$ is of the form $2^{n+R}+Sn^2+Tn+U$ for all $n>0.$ Find $R,S,T,$ and $U.$
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 \( \boxed{ R = 1, S = 0, T = -1, U = -2 } \).