Write the recursive and explicit formulas for an arithmetic sequence with a1 = 3 and d = 4?
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An arithmetic sequence is a sequence of numbers where the difference between consecutive terms is constant. In this example, we're given the first term a sub 1 equals 3 and the common difference d equals 4. This generates the sequence 3, 7, 11, 15, 19, and so on.
The common difference d equals 4 means that each term is exactly 4 more than the previous term. We can see this pattern clearly: starting with 3, we add 4 to get 7, then add 4 again to get 11, and so on. This constant addition is what defines an arithmetic sequence.
The recursive formula for an arithmetic sequence expresses each term in relation to the previous term. The general form is a sub n equals a sub n minus one plus d. For our sequence with d equals 4, this becomes a sub n equals a sub n minus one plus 4. With the initial condition a sub 1 equals 3, we can calculate subsequent terms: a sub 2 equals 3 plus 4 equals 7, a sub 3 equals 7 plus 4 equals 11, and a sub 4 equals 11 plus 4 equals 15.
While the recursive formula is useful, we can also develop an explicit formula that directly calculates any term based on its position. Let's examine the pattern: a sub 2 equals a sub 1 plus 1 times d, a sub 3 equals a sub 1 plus 2 times d, a sub 4 equals a sub 1 plus 3 times d. We notice that the multiplier of d is always one less than the term number. This leads us to the general explicit formula: a sub n equals a sub 1 plus (n minus 1) times d.
Now let's derive the explicit formula for our specific sequence. We start with the general formula: a sub n equals a sub 1 plus (n minus 1) times d. Substituting our values a sub 1 equals 3 and d equals 4, we get a sub n equals 3 plus (n minus 1) times 4. Distributing the 4 gives us a sub n equals 3 plus 4n minus 4. Combining like terms, we arrive at our final explicit formula: a sub n equals 4n minus 1. Let's verify this works: a sub 1 equals 4 times 1 minus 1 equals 3, a sub 2 equals 4 times 2 minus 1 equals 7, and so on.
Let's verify that both our recursive and explicit formulas produce the same results. We'll find the 5th, 10th, and 20th terms using both methods. For the recursive method, we build up from the previous terms: a sub 5 equals a sub 4 plus 4 equals 19, a sub 10 equals a sub 9 plus 4 equals 39, and a sub 20 equals a sub 19 plus 4 equals 79. Using the explicit formula a sub n equals 4n minus 1: a sub 5 equals 4 times 5 minus 1 equals 19, a sub 10 equals 4 times 10 minus 1 equals 39, and a sub 20 equals 4 times 20 minus 1 equals 79. Both methods give identical results, confirming our formulas are correct.
Each formula has its practical applications. The explicit formula is ideal when we need to find a distant term directly. For example, to find the 50th term, we simply substitute into our formula: a sub 50 equals 4 times 50 minus 1 equals 199. This is much more efficient than calculating all 49 previous terms. The recursive formula is most useful when we know a specific term and need the next one. If we're given that a sub 15 equals 59, we can immediately find a sub 16 equals a sub 15 plus 4 equals 63. The comparison table summarizes when to use each approach: recursive formulas are best for consecutive terms, while explicit formulas are best for distant terms. In terms of efficiency, recursive formulas become inefficient for distant terms, while explicit formulas maintain high efficiency for any term.
Let's summarize our complete solution. We have two formulas for our arithmetic sequence. The recursive formula is a sub n equals a sub n minus one plus 4, with the initial condition a sub 1 equals 3. The explicit formula is a sub n equals 4n minus 1. These formulas represent two different ways of thinking about the sequence: the recursive formula shows the relationship between consecutive terms, while the explicit formula allows us to calculate any term directly by its position. Both formulas generate the same sequence: 3, 7, 11, 15, 19, 23, and so on. Understanding both approaches gives us flexibility in solving various problems involving arithmetic sequences.