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GCSE :: Quadratic Sequences

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1 GCSE :: Quadratic Sequences
Dr J Frost Objectives: Determine the nth term formula for sequences with a constant second difference. Last modified: 27th December 2018

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3 RECAP: What do we mean by 𝑛th term?
An 𝑛th term formula is an expression in terms of 𝑛 that allows us to find the value at a particular position in the sequence. For example, we can find the 3rd term in the sequence by just making 𝑛=3. What is the 5th term in the sequence with 𝑛th term: 3𝑛 →𝟑 𝟓 +𝟕=𝟐𝟐 𝑛 2 +2𝑛 → 𝟓 𝟐 +𝟐 𝟓 =𝟑𝟓 ? ? Suppose we found the first 5 terms in a sequence where the 𝑛th term formula is a quadratic expression (i.e. has a 𝑛 2 in it*): Find the first 5 terms of the sequence with 𝑛th term 𝑛 2 +3𝑛−1. What do you notice? 1st term: 𝟏 𝟐 +𝟑 𝟏 −𝟏=𝟑 First five terms: 𝟑, 𝟗, 𝟏𝟕, 𝟐𝟕, 𝟑𝟗 Notice that the “difference of the difference”, i.e. the second difference, stays the same. We will see how we can use this to work out the formula if we had the sequence. ? +6 +8 +10 +12 +2 +2 +2 * More formally, is of the form 𝑎 𝑛 2 +𝑏𝑛+𝑐 where 𝑎,𝑏,𝑐 are constants and 𝑎 is not 0).

4 RECAP: Arithmetic Sequences
An arithmetic sequence is one where the difference between terms is the same. Recall that we make this difference the coefficient of (i.e. the number in front of) the 𝑛 term, e.g. Find the 𝑛th term formula for the arithmetic sequence starting: 2, 5, 8, 11, 14, … ? Thinking process: The difference between terms is 3 so we slap this on front of an 𝒏 and start our formula with 𝟑𝒏. If the formula was 𝟑𝒏, our first term would be 𝟑×𝟏=𝟑 But the first term should be 2, so we need to subtract 1 from our formula. The 𝒏th term formula is therefore 𝟑𝒏−𝟏 We’re basically using an approach of starting the formula with something sensible (here 3𝑛), seeing what sequence this would give us, and ‘adjusting’ appropriately. This will be important when we get to quadratic sequences.

5 Quickfire Examples Find formulae for the 𝑛th term of each of these sequences: (in your head) ? 4, 7, 10, 13, 16, … 7, 12, 17, 22, 27, … 3, 10, 17, 24, 31, … 12, 9, 6, 3, 0, −3, … →3𝑛+1 →5𝑛+2 →7𝑛−4 →15−3𝑛 ? ? ?

6 Working out the 𝑛th term
Find the 𝑛th term formula for the sequence starting 3, 8, 15,24,35,… STEP 1: Write out 𝑛 and the original sequence. 𝑛 𝑛th term STEP 2: Work out second difference. +5 +7 +9 +2 STEP 3: Halve this to find coefficient of 𝑛 2 term. 1𝑛 2 STEP 4: Work out what we need to add to get from this to correct term. Work out its formula. To add Difference is 2 so formula for this row is 𝟐𝒏 +2 +2 The 𝑛th term is the sum of this initial sequence (the 1 𝑛 2 ) and the adjustment (the 2𝑛), giving 𝒏 𝟐 +𝟐𝒏

7 Further Example Find the 𝑛th term formula for the sequence starting 4, 15, 32, 55, 84,… STEP 1: Write out 𝑛 and the original sequence. 𝑛 𝑛th term STEP 2: Work out second difference. +11 +17 +23 +6 STEP 3: Halve this to find coefficient of 𝑛 2 term. Beware: 3 𝑛 2 means 3× 𝑛 2 and NOT 3𝑛 2 3𝑛 2 STEP 4: Work out what we need to add to get from this to correct term. Work out its formula. To add Formula for this adjustment row is 𝟐𝒏−𝟏 +2 +2 𝒏th term =𝟑 𝒏 𝟐 +𝟐𝒏−𝟏

8 Quickfire First Step Don’t get mixed up: If the first difference is constant, we leave this difference as it is and shove it on front of 𝒏 to start our formula. If the second difference is constant, we halve this difference and shove it on front of 𝒏 𝟐 to start our formula. How do we start the formula? How do we start the formula? 4, 6, 8, 10, 12, … a 𝟐𝒏 ? 5, 9, 13, 17, 21, … ? e 𝟒𝒏 +2 +2 +2 +4 +4 +4 1, 2, 7, 16, 25, … ? 𝟐 𝒏 𝟐 b 1, 4, 9, 16, 25, … 𝒏 𝟐 ? f +1 +5 +9 +3 +5 +7 +4 +4 ? +2 +2 5, 13, 21, 29, 37, … 𝟖𝒏 c 7, 6, 5, 4, 3, … −𝒏 ? g +8 +8 +8 −1 −1 −1 ? 3, 5, 13, 27, 47, … 𝟑 𝒏 𝟐 d 0, 0, 10, 30, 60, … 𝟓 𝒏 𝟐 ? h +2 +8 +14 +0 +10 +20 +6 +6 +10 +10

9 Test Your Understanding
Determine the full 𝑛th term formula for the following sequences. 𝒏th term: 3, 9, 19, 33, 51, … 4, 10, 18, 28, … 4, 9, 18, 31, 48, … 2 𝑛 2 +1 ? 𝑛 2 +3𝑛 ? ? 4 𝑛 2 −𝑛+3 Fro Tip: If a GCSE question wants to be on the harder side, they’ll make the 𝑛 term negative (or possibly the 𝑛 2 term fractional)

10 Exercise 𝟐 𝒏 𝟐 𝒏 𝟐 +𝒏+𝟏 𝟐 𝒏 𝟐 +𝟐𝒏+𝟏 𝒏 𝟐 −𝟐𝒏+𝟑 𝟑 𝒏 𝟐 +𝟑𝒏−𝟏 𝟏 𝟐 𝒏 𝟐 + 𝟏 𝟐 𝒏 𝟐 𝒏 𝟐 +𝟑𝒏−𝟒 − 𝒏 𝟐 +𝟒𝒏+𝟏 𝟏 𝟐 𝒏 𝟐 +𝟓𝒏−𝟏 2, 8, 18, 32, … 3, 7, 13, 21, 31, … 5, 13, 25, 41, 61, 85, … 2, 3, 6, 11, 18, … 5, 17, 35, 59, 89, 125, … 1, 3, 6, 10, 15, 21, … 1, 10, 23, 40, 61, 86, … 4, 5, 4, 1, −4, −11, … 3.5, 7, 9.5, 11, 11.5, 11, … ? 1 ? 2 ? 3 ? 4 ? These are the triangular numbers. ? 5 6 ? 7 ? ? 8 In general, the 𝑛th term formula for a quadratic sequence is 𝑎 𝑛 2 +𝑏𝑛+𝑐. Find the difference between the 𝑛th and (𝑛+1)th terms in terms of 𝑎,𝑏,𝑐 and 𝑛, and between the (𝑛+1)th and 𝑛+2 th terms, and hence the second difference. N The second difference is 𝟐𝒂 (and notice that halving this gives the 𝒂 on front of the 𝒏 𝟐 term). This is discussed in the next few slides… ?

11 Exit GCSE Past Paper Question!
So, do you get it? ? 𝟐 𝒏 𝟐 +𝒏+𝟏

12 Just For Your Interest…
Why do we halve the second difference for the first term of a quadratic sequence? Let’s first consider arithmetic sequences (where the first difference is constant). We know the general form for the 𝑛th term formula is 𝑎𝑛+𝑏: Current Term Next Term 𝑛 ? 𝑛 + 1 Position Term 𝑎𝑛 + 𝑏 𝑎𝑛 + 𝑎 + 𝑏 ? +𝑎 ? We can see the difference between terms is 𝑎, but is also the number in front of the 𝑛 term. This justifies why we shove the first difference in front of the 𝑛.

13 Just For Your Interest…
Current Term Next Term Term after that ? 𝑛 ? 𝑛+1 𝑛 + 2 Position Term 𝑎𝑛2+𝑏𝑛+𝑐 𝑎𝑛2+2𝑎𝑛+𝑏𝑛+𝑎+𝑏+𝑐 ? 𝑎𝑛2+4𝑎𝑛 +𝑏𝑛+4𝑎+2𝑏+𝑐 ? ? ? 2𝑎𝑛+𝑎+𝑏 2𝑎𝑛+3𝑎+𝑏 2𝑎 ? Since the second difference is 2𝑎 and the coefficient of 𝑛 2 is 𝑎, we can see halving the second difference gives us the coefficient of 𝑛 2 .

14 Finding a formula using simultaneous equations
There is an alternative method of finding the 𝑛th term formula using simultaneous equations. (Note that we often use 𝑢 𝑛 to denote the 𝑛th term, so 𝑢 1 would be the first term) You’re given the first three terms of a quadratic (second difference) sequence: 𝒖𝟏=𝟑, 𝒖𝟐=𝟕, 𝒖𝟑=𝟏𝟓 We know that we can use: 𝒖 𝒏 =𝒂 𝒏 𝟐 +𝒃𝒏+𝒄 What equations can we form? ? 𝒂+𝒃+𝒄 =𝟑 (𝟏) 𝟒𝒂+𝟐𝒃+𝒄=𝟕 (𝟐) 𝟗𝒂+𝟑𝒃+𝒄 = 𝟏𝟓 (𝟑) Solve by elimination: 2 − 1 : 3𝑎+𝑏=4 3 − 1 : 8𝑎+2𝑏=12 𝑎=2, 𝑏=−2, 𝑐=3 𝑢 𝑛 =2 𝑛 2 −2𝑛+3 ?

15 1,4,13,… Test Your Understanding and Extension ? ? ? ? ?
𝑎+𝑏+𝑐=1 4𝑎+2𝑏+𝑐=4 9𝑎+3𝑏+𝑐=13 ? ? 1,4,13,… 𝑢 𝑛 =3 𝑛 2 −6𝑛+4 Oxford Maths Admissions Exam If 𝑥 𝑛 refers to the 𝑛th term, then 𝑥 𝑛−1 refers to the previous term. 𝑥4 = 10, 𝑥5 = 15 ? 𝐴=0, 𝐵=0.5, 𝐶=0.5 ? 𝑛 = 40 ?


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