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Quadratic Sequences Simplified Revision Notes

Revision notes with simplified explanations to understand Quadratic Sequences quickly and effectively.

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Quadratic Sequences

A Quadratic Sequence is a type of sequence where the difference between the terms isn't always the same, but if you look at the differences between those differences, they will be the same. This may sound a bit confusing at first, so let's break it down.

For example:

  • Consider the sequence: 4,7,12,19,28,...4, 7, 12, 19, 28, ...
    • The first differences (how much the sequence goes up each time) are: 3,5,7,93, 5, 7, 9 (these aren't the same, so it's not a simple pattern).
    • The second differences (the differences between the first differences) are: 2,2,22, 2, 2 (these are the same, which tells us this is a quadratic sequence).

Finding the Formula for a Quadratic Sequence

To find a formula (or general term) that tells you what any term in the sequence will be, we use this formula: Tn=an2+bn+cT_n = an^2 + bn + c where:

  • TnT_n is the term you're trying to find (like the 1st1st term, 2nd2nd term, etc.).
  • nn is the position of that term in the sequence.
  • aa, bb, and cc are numbers we need to figure out by using the sequence.

Steps to Find the General Term:

  1. Find the Second Difference:
  • Look at the differences between the terms, then look at the differences between those differences. The second difference helps you figure out the value of aa. Specifically, 2a2a equals the second difference, so divide the second difference by 22 to find aa.
infoNote

Exam Tip: It can be helpful to remember that aa is half the second difference.

  1. Create Two Equations:
  • Use the first two terms of the sequence to create two equations. These equations help you find the values of bb and cc.
  1. Solve for bb and cc:
  • Once you have aa, plug it into your equations and solve for bb and cc.
  1. Write the Formula:
  • Now that you have aa, bb, and cc, you can write the formula for the sequence.
infoNote

Worked Example:

Let's find the formula for the sequence 4,7,12,19,28,...4, 7, 12, 19, 28, ...


  1. Find the First and Second Differences: Term number12345Sequence47121928First difference3579Second difference222\begin{array}{c|c|c|c|c} \text{Term number} & 1 & 2 & 3 & 4 & 5 \\ \hline \text{Sequence} & 4 & 7 & 12 & 19 & 28 \\ \hline \text{First difference} & & 3 & 5 & 7 & 9 \\ \hline \text{Second difference} & & & 2 & 2 & 2 \\ \end{array}
  • The second difference is 22, so 2a=22a = 2, which means a=1a = 1.

  1. Create Two Equations Using the First Two Terms:
  • The general term is Tn=an2+bn+cT_n = an^2 + bn + c.

  • Substitute a=1a = 1 into the formula: Tn=n2+bn+cT_n = n^2 + bn + c. Now, use the first two terms in the sequence to make two equations.

  • For the 1st term (n=1)( n = 1 ), we know T1=4T_1 = 4: 12+b(1)+c=41+b+c=4b+c=3(Equation 1)1^2 + b(1) + c = 4 \quad \Rightarrow \quad 1 + b + c = 4 \quad \Rightarrow \quad b + c = 3 \quad \text{(Equation 1)}

  • For the 2nd2nd term (n=2)( n = 2 ), we know T2=7T_2 = 7: 22+b(2)+c=74+2b+c=72b+c=3(Equation 2)2^2 + b(2) + c = 7 \quad \Rightarrow \quad 4 + 2b + c = 7 \quad \Rightarrow \quad 2b + c = 3 \quad \text{(Equation 2)}


  1. Solve the Equations:
  • Subtract Equation 11 from Equation 22 to find bb: (2b+c)(b+c)=33:success[b=0](2b + c) - (b + c) = 3 - 3 \quad \Rightarrow \quad :success[b = 0]
  • Substitute b=0b = 0 back into Equation 11 to find cc: b+c=30+c=3:success[c=3]b + c = 3 \quad \Rightarrow \quad 0 + c = 3 \quad \Rightarrow \quad :success[c = 3]
  • Now, we have a=1a = 1, b=0b = 0, and c=3c = 3.

  1. Write the Formula:
  • Substitute aa, bb, and cc into the general term formula: :highlight[Tn=n2+3]:highlight[T_n = n^2 + 3]
  • This is the formula for the sequence.

Checking the Formula:

Let's check if the formula works by calculating a few terms:

  • For the 1st1st term (n=1)( n = 1 ): T1=12+3=1+3=4:success[(Correct!)]T_1 = 1^2 + 3 = 1 + 3 = 4 \quad :success[\text{(Correct!)}]
  • For the 2nd2nd term (n=2)( n = 2 ): T2=22+3=4+3=7:success[(Correct!)]T_2 = 2^2 + 3 = 4 + 3 = 7 \quad :success[\text{(Correct!)}]
  • For the 3rd3rd term (n=3)( n = 3 ): T3=32+3=9+3=12:success[(Correct!)]T_3 = 3^2 + 3 = 9 + 3 = 12 \quad :success[\text{(Correct!)}] The formula Tn=n2+3T_n = n^2 + 3 correctly gives us the terms in the sequence.
infoNote

Exam Tip:

  • Always double-check your formula by substituting it back into the sequence to make sure it gives the right terms.
  • Practice with different quadratic sequences to get comfortable with finding aa, bb, and cc. By following these steps, you'll be ready to tackle quadratic sequences on your Junior Cycle Maths exam!

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