Math Problem Statement

Here are the first 5 terms of a quadratic sequence: 1, 3, 7, 13, 21. Find an expression, in terms of n, for the nth term of this quadratic sequence.

Solution

The sequence provided is: 1,3,7,13,211, 3, 7, 13, 21

To find the general formula for the nn-th term of a quadratic sequence, we need to determine the quadratic expression, typically written as: an=an2+bn+ca_n = an^2 + bn + c where aa, bb, and cc are constants to be determined.

Step 1: Find the first and second differences

  1. First differences (difference between consecutive terms): 31=2,73=4,137=6,2113=83 - 1 = 2, \quad 7 - 3 = 4, \quad 13 - 7 = 6, \quad 21 - 13 = 8 So, the first differences are: 2,4,6,82, 4, 6, 8

  2. Second differences (difference between consecutive first differences): 42=2,64=2,86=24 - 2 = 2, \quad 6 - 4 = 2, \quad 8 - 6 = 2 The second differences are constant, which confirms that this is a quadratic sequence.

Step 2: Use the second difference to find aa

For a quadratic sequence, the second difference is always equal to 2a2a. Therefore: 2a=2a=12a = 2 \quad \Rightarrow \quad a = 1

Step 3: Set up the general formula

Now we know that the general term has the form: an=n2+bn+ca_n = n^2 + bn + c

Step 4: Use known terms to solve for bb and cc

We can substitute values from the given sequence to form a system of equations. Using the first three terms of the sequence:

  1. For n=1n = 1: a1=12+b(1)+c=1a_1 = 1^2 + b(1) + c = 1, which simplifies to: 1+b+c=1b+c=0(Equation 1)1 + b + c = 1 \quad \Rightarrow \quad b + c = 0 \quad \text{(Equation 1)}

  2. For n=2n = 2: a2=22+b(2)+c=3a_2 = 2^2 + b(2) + c = 3, which simplifies to: 4+2b+c=32b+c=1(Equation 2)4 + 2b + c = 3 \quad \Rightarrow \quad 2b + c = -1 \quad \text{(Equation 2)}

Step 5: Solve the system of equations

From Equation 1, b+c=0b + c = 0, so c=bc = -b. Substituting this into Equation 2: 2bb=1b=12b - b = -1 \quad \Rightarrow \quad b = -1 Substitute b=1b = -1 back into Equation 1: 1+c=0c=1-1 + c = 0 \quad \Rightarrow \quad c = 1

Step 6: Final expression

Substitute a=1a = 1, b=1b = -1, and c=1c = 1 into the general form: an=n2n+1a_n = n^2 - n + 1

Thus, the expression for the nn-th term of the sequence is: an=n2n+1a_n = n^2 - n + 1

Would you like further details on the process, or do you have any other questions?

Related Questions:

  1. How can I verify that an=n2n+1a_n = n^2 - n + 1 works for other terms in the sequence?
  2. What is the difference between quadratic sequences and arithmetic sequences?
  3. Can quadratic sequences be used to model real-world phenomena?
  4. How would the formula change if the second differences were not constant?
  5. What happens if I add a constant to a quadratic sequence—how does that affect the formula?

Tip:

Quadratic sequences always have a constant second difference, which is key to recognizing them and deriving their formulas!

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Math Problem Analysis

Mathematical Concepts

Quadratic Sequences
Algebra
Sequences and Series

Formulas

n^2 + bn + c

Theorems

Quadratic Formula for Sequences

Suitable Grade Level

Grades 9-10