Recursive and Explicit Formulas for Arithmetic Sequences

Gives the 1st term a(1) and a rule for finding the nth term represented as a(n) given the previous term which is the (n-1)th term represented as a(n-1).

• The recursive formula for a sequence gives the 1st term represented by a(1) and a rule for finding the nth term represented by a(n) given the previous term which is the (n-1)th term represented by a(n-1).

• The drawback is needing all previous terms to find the nth term.

• Any arithmetic sequence can be represented by:

  a(1), a(2), a(3), . . . , a(n-1), a(n), a(n+1), . . .

Gives the 1st term a(1) and a rule for finding the nth term represented as a(n) given the previous term which is the (n-1)th term represented as a(n-1).

• Example of recursive formula for the arithmetic sequence -1, 4, 9, 14, 19, . . .

• Recursive formula: a(1) = -1, a(n) = a(n-1) + 5

• The 1st term a(1) is -1 and the common difference d is 5, so we show that the nth term a(n) is equal to the previous term a(n-1) + 5. This means we add 5 to the previous term to get the next term.

• The 6th term a(6) = a(6-1) + 5 = a(5) + 5 = 19 + 5 = 24.

Gives the nth term without having to find all previous terms.
a(n) = a(1) + (n-1)(d), where a(1) is the 1st term and d is the common difference found by subtracting any term from the term which follows it.

• The explicit formula allows finding the nth term represented as a(n) without having to find all previous terms.

• The explicit formula can be written as a(n) = a(1) + (n-1)(d), where a(n) is the nth term, a(1) is the 1st term, n is the term number, and d is the common difference for the arithmetic sequence which is either given or can be calculated by subtracting any term from the term which follows it.

• For example, d = a(2) - a(1) = 2nd term - 1st term.

​Gives the nth term without having to find all previous terms.
a(n) = a(1) + (n-1)(d), where a(1) is the 1st term and d is the common difference found by subtracting any term from the term which follows it.

• Example of explicit formula for the arithmetic sequence

  -1, 4, 9, 14, 19, . . .

• Explicit formula: a(n) = a(1) + (n-1)(d), so

  a(n) = -1 + (n-1)(5) since a(1) = -1 and d = 4 - (-1) = 5.

• The 6th term a(6) = -1+(6-1)(5) = -1+(5)(5) = -1+25 = 24.

Recursive Formula: a(1) = #, a(n) = a(n-1) + d
Explicit Formula: a(n) = a(1) + (n-1)(d)

• Given the arithmetic sequence: 3, -1, -5, -9, . . .

• We have the 1st term a(1) = 3 and the common difference d = -1 - 3 = -4 which means we add -4 (or subtract 4) to get the next term in the sequence. We also have a(2) = -1, a(3) = -5, a(4) = -9.

• The recursive formula is: a(1) = 3, a(n) = a(n-1) + -4 which can also be written as a(n) = a(n-1) - 4.

• The explicit formula is: a(n) = 3 + (n-1)(-4) which can be simplified using the distributive property to get a(n) = 3 + -4n + 4 which is a(n) = 7 - 4n.

​Recursive Formula: a(1) = #, a(n) = a(n-1) + d
Explicit Formula: a(n) = a(1) + (n-1)(d)

• Given the arithmetic sequence: 3, -1, -5, -9, . . .

• To find the 5th term a(5), we can use either formula since we have a(4) = -9 and the recursive formula would give a(5) = a(5-1) - 4 = a(4) - 4 = -9 - 4 = -13, so a(5) = -13 using the recursive formula.

• The explicit formula gives a(5) = 3 + (5-1)(-4) = 3 + (4)(-4) = 3 + -16 = -13, so a(5) = -13 using the explicit formula.

• We can check the simplified explicit formula a(n) = 7 - 4n as well to show a(5) = 7 - 4(5) = 7 - 20 = -13, so a(5) = -13.