Consider the sequence defined by the recursive formula \( a_n = 2a_{n-1} - 3 \) with the initial term \( a_1 = 4 \). Find the explicit formula for the \( n \)-th term of the sequence.

["Understanding the Recursive Sequence: An Explicit Formula for ( a_n = 2a_{n-1} - 3 ) with ( a_1 = 4 )", "Recursive sequences are a fundamental concept in mathematics, offering valuable insights into patterns and relationships over time. One such sequence is defined by the recursive formula:\n[\na_n = 2a_{n-1} - 3\n]\nwith the initial term ( a_1 = 4 ). While recursion defines each term in terms of the previous one, finding a direct formula for the ( n )-th term—known as the explicit formula—allows efficient calculation without computing all prior terms.", "In this article, we explore how to derive the explicit expression for ( a_n ), demystifying the recursive sequence and revealing its growth behavior.", "---", "### Step 1: Recognizing the Type of Recurrence", "The given recurrence is linear and nonhomogeneous due to the constant term ( -3 ). The general form of such sequences is:\n[\na_n = c \cdot a_{n-1} + d\n]\nwhere ( c = 2 ) and ( d = -3 ). Linear nonhomogeneous recursions have solutions composed of two parts:\n1. The homogeneous solution solving ( a_n^{(h)} = c \cdot a_{n-1}^{(h)} )\n2. A particular solution that accounts for the constant term ( d )", "---", "### Step 2: Solve the Homogeneous Equation", "The homogeneous version of the recurrence is:\n[\na_n^{(h)} = 2a_{n-1}^{(h)}\n]\nThis is a simple geometric sequence with solution:\n[\na_n^{(h)} = A \cdot 2^n\n]\nfor some constant ( A ).", "---", "### Step 3: Find a Particular Solution", "Since the nonhomogeneous term is constant (( d = -3 )), we assume a constant particular solution:\n[\na_n^{(p)} = C\n]\nSubstitute into the original recurrence:\n[\nC = 2C - 3\n]\nSolving for ( C ):\n[\nC - 2C = -3 \Rightarrow -C = -3 \Rightarrow C = 3\n]\nThus, the particular solution is ( a_n^{(p)} = 3 ).", "---", "### Step 4: Combine Solutions", "The general solution is the sum:\n[\na_n = a_n^{(h)} + a_n^{(p)} = A \cdot 2^n + 3\n]", "---", "### Step 5: Use Initial Condition to Solve for ( A )", "Apply the initial condition ( a_1 = 4 ):\n[\na_1 = A \cdot 2^1 + 3 = 2A + 3 = 4\n]\nSolve:\n[\n2A = 1 \Rightarrow A = \frac{1}{2}\n]", "---", "### Step 6: Write the Final Explicit Formula", "Substitute ( A = \frac{1}{2} ) back into the general solution:\n[\na_n = \frac{1}{2} \cdot 2^n + 3 = 2^{n-1} + 3\n]", "---", "### Final Answer", "The explicit formula for the ( n )-th term of the sequence is:\n[\n\boxed{a_n = 2^{n-1} + 3}\n]", "---", "### Interpretation and Use", "This closed-form expression allows direct computation of any term:\n- For ( n = 1 ): ( a_1 = 2^0 + 3 = 4 ) ✅\n- For ( n = 2 ): ( a_2 = 2^1 + 3 = 5 ), checking: ( a_2 = 2(4) - 3 = 8 - 3 = 5 ) ✅\n- For ( n = 3 ): ( a_3 = 2^2 + 3 = 7 ), checking: ( a_3 = 2(5) - 3 = 10 - 3 = 7 ) ✅", "The explicit formula reveals the sequence grows exponentially (due to ( 2^{n-1} )), with a constant offset of 3.", "Understanding recursive sequences through their explicit forms empowers problem-solving in discrete mathematics, algorithm analysis, and applied modeling—making this derivation a valuable skill for students and practitioners alike."]