$024 a_{1}=3, a_{n+1}-a_{n}=2^{n}$

Step 1 ：Given the first term of the sequence, \(a_1 = 3\), and a recursive formula for the sequence, \(a_{n+1} - a_n = 2^n\).

Step 2 ：Rewrite the recursive formula as \(a_{n+1} = a_n + 2^n\).

Step 3 ：Expand this formula for a few terms to find a pattern:

Step 4 ：For \(n = 1\), we have \(a_2 = a_1 + 2^1 = 3 + 2 = 5\).

Step 5 ：For \(n = 2\), we have \(a_3 = a_2 + 2^2 = 5 + 4 = 9\).

Step 6 ：For \(n = 3\), we have \(a_4 = a_3 + 2^3 = 9 + 8 = 17\).

Step 7 ：Find a general formula for the sequence by summing the powers of 2 up to \(n\) and adding the first term: \(a_n = a_1 + \sum_{i=1}^{n-1} 2^i\).

Step 8 ：Calculate the first few terms of the sequence using the general formula: \(3, 5, 9, 17, 33\).

Step 9 ：\(\boxed{\text{The first few terms of the sequence are 3, 5, 9, 17, and 33. The general formula for the sequence is } a_n = a_1 + \sum_{i=1}^{n-1} 2^i\text{.}}\)