Math Problem Statement

Solve this

The start of a quadratic sequence is shown below. What is the nth term rule for the sequence? -6, -3, 2, 9, 18, ...

Solution

This problem involves finding the general nn-th term of a quadratic sequence. Let’s solve it step by step.

Sequence: 6,3,2,9,18,-6, -3, 2, 9, 18, \dots

Step 1: Find the first differences.

First differences: (3(6)),(2(3)),(92),(189)\text{First differences: } (-3 - (-6)), (2 - (-3)), (9 - 2), (18 - 9) =3,5,7,9= 3, 5, 7, 9

Step 2: Find the second differences.

Second differences: (53),(75),(97)\text{Second differences: } (5 - 3), (7 - 5), (9 - 7) =2,2,2= 2, 2, 2

Since the second differences are constant, this confirms the sequence is quadratic.

Step 3: General form of a quadratic sequence.

The nn-th term of a quadratic sequence is given by: Tn=an2+bn+cT_n = an^2 + bn + c We need to determine the coefficients aa, bb, and cc by substituting values for nn and TnT_n.

Step 4: Solve for aa, bb, and cc.

Using the terms:

  • For n=1,T1=6n = 1, T_1 = -6: a(1)2+b(1)+c=6a+b+c=6a(1)^2 + b(1) + c = -6 \quad \Rightarrow \quad a + b + c = -6

  • For n=2,T2=3n = 2, T_2 = -3: a(2)2+b(2)+c=34a+2b+c=3a(2)^2 + b(2) + c = -3 \quad \Rightarrow \quad 4a + 2b + c = -3

  • For n=3,T3=2n = 3, T_3 = 2: a(3)2+b(3)+c=29a+3b+c=2a(3)^2 + b(3) + c = 2 \quad \Rightarrow \quad 9a + 3b + c = 2

This gives us the system of equations:

  1. a+b+c=6a + b + c = -6
  2. 4a+2b+c=34a + 2b + c = -3
  3. 9a+3b+c=29a + 3b + c = 2

Step 5: Solve the system of equations.

  • Subtract (1) from (2): (4a+2b+c)(a+b+c)=3(6)(4a + 2b + c) - (a + b + c) = -3 - (-6) 3a+b=3(Equation 4)3a + b = 3 \quad \text{(Equation 4)}

  • Subtract (2) from (3): (9a+3b+c)(4a+2b+c)=2(3)(9a + 3b + c) - (4a + 2b + c) = 2 - (-3) 5a+b=5(Equation 5)5a + b = 5 \quad \text{(Equation 5)}

  • Subtract (4) from (5): (5a+b)(3a+b)=53(5a + b) - (3a + b) = 5 - 3 2a=2a=12a = 2 \quad \Rightarrow \quad a = 1

  • Substitute a=1a = 1 into (4): 3(1)+b=3b=03(1) + b = 3 \quad \Rightarrow \quad b = 0

  • Substitute a=1a = 1 and b=0b = 0 into (1): 1+0+c=6c=71 + 0 + c = -6 \quad \Rightarrow \quad c = -7

Step 6: Write the nn-th term formula.

Tn=an2+bn+cTn=n27T_n = an^2 + bn + c \quad \Rightarrow \quad T_n = n^2 - 7

Final Answer:

The nn-th term rule for the sequence is: Tn=n27T_n = n^2 - 7

Let me know if you'd like further clarifications! Here are some related questions for practice:


  1. How would you check if a given sequence is quadratic?
  2. Can you determine the nn-th term formula for the sequence 2,6,12,20,30,2, 6, 12, 20, 30, \dots?
  3. What happens to the nn-th term formula if the second differences are not constant?
  4. How can you verify your answer for the nn-th term using substitution?
  5. What is the difference between an arithmetic sequence and a quadratic sequence?

Tip: For quadratic sequences, always confirm the second differences are constant before proceeding.

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Math Problem Analysis

Mathematical Concepts

Sequences
Quadratic Sequences
Algebra

Formulas

General form of a quadratic sequence: Tn = an^2 + bn + c
Finding first and second differences
System of equations to solve coefficients a, b, c

Theorems

A sequence with constant second differences is quadratic

Suitable Grade Level

Grades 8-10