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10.1 Constant Differences
In the first part of this unit we consider sequences where the difference between successive terms is the same every time. We also use formulae to create the terms of a sequence.
Example 1
Write down the next 3 terms of each of the following sequences: (a) 7, 11, 15, 19, 23, ... (b) 1, 9, 17, 25, 33, ...
Solution
(a) 7 11 15 19 23 ...
4 4 4 4 The difference between each term and the next is always 4. This value is called the first difference. So we can continue the sequence by adding 4 each time. This gives the sequence: 7, 11, 15, 19, 23, 27, 31, 35 (b) 1 9 17 25 33 ...
8 8 8 8 Here the difference between each term and the next is always 8. To continue the sequence we must keep on adding 8 every time. This gives the sequence: 1, 9, 17, 25, 33, 41, 49, 57
Example 2
A sequence is defined by the formula u (^) n = 3 n + 1. Calculate the first 5 terms of this sequence.
Solution
The first term, often called u (^) 1 , is formed by substituting n = 1 into the formula. u (^) 1 = 3 × 1 + 1 = 3 + 1 = 4
10 Sequences
For the second term, substitute n^ =^2 to give:
u (^) 2 = 3 × 2 + 1 = 7
For the third term, substitute n = 3 to give:
u (^) 3 = 3 × 3 + 1 = 10
For the fourth term, substitute n = 4 to give:
u (^) 4 = 3 × 4 + 1 = 13
For the fifth term, substitute n = 5 to give:
u (^) 5 = 3 × 5 + 1 = 16
So the first 5 terms of the sequence are 4, 7, 10, 13, 16.
Example 3
The terms of a sequence are given by the formula u (^) n = 8 n − 3.
Calculate: (a) the first 3 terms of the sequence, (b) the 100th term of the sequence, (c) the 200th term of the sequence.
Solution
(a) n = 1 gives u (^) 1 = 8 × 1 − 3 = 5 n = 2 gives u (^) 2 = 8 × 2 − 3 = 13 n = 3 gives u (^) 3 = 8 × 3 − 3 = 21 So the first 3 terms are 5, 13, 21.
- A sequence is given by u (^) n = 9 n + 2.
(a) Calculate the first 4 terms of the sequence. (b) How does the difference between terms relate to the formula?
- A sequence is given by the formula u (^) n = 11 n − 7.
(a) What would you expect to be the difference between the terms of the sequence? (b) Calculate the first 4 terms of the sequence and check your answer to part (a). (c) Calculate the 10th term of the sequence.
- A sequence is defined by the formula u (^) n = 82 − 4 n.
(a) Calculate the first 5 terms of the sequence. (b) What is the difference between terms for the sequence? (c) How does this difference relate to the formula? (d) Calculate the 20th term of the sequence.
- (a) Calculate the 100th term of the sequence given by u (^) n = 8 n − 5.
(b) Calculate the 25th term of the sequence given by u (^) n = 11 n − 3. (c) Calculate the 200th term of the sequence given by u (^) n = 3 n + 22. (d) Calculate the 58th term of the sequence defined by u^ n =^1000 −^5 n.
- Four sequences, A, B, C and D, are defined by the following formulae:
A u (^) n = 8 n + 2 B u (^) n = 7 n − 3 C u (^) n = 3 n + 1 D u (^) n = 100 − 6 n (a) Which sequences have 4 as their first term? (b) Which sequence is decreasing? (c) Which sequence has a difference of 7 between terms? (d) Which sequence has 301 as its 100th term?
- (a) Look at this part of a number line. Write down the 2 missing numbers.
- 7 ......... 1 5 9 ......... 17
Copy and complete this sentence: The numbers on this line go up in steps of .......... (b) This is a different number line. Write down the 3 missing numbers. 7.5 7.6 7.7 7. ......... ......... .........
Copy and complete this sentence: The numbers on this line go up in steps of .......... (KS3/97/Ma/Tier 4-6/P1)
- Jeff makes a sequence of patterns with black and grey triangular tiles.
pattern number 1
pattern number 2
pattern number 3 The rule for finding the number of tiles in pattern number N in Jeff's sequence is: number of tiles = 1 +3N (a) The 1 in this rule represents the black tile. What does the 3N represent? (b) Jeff makes pattern number 12 in his sequence. How many black tiles and how many grey tiles does he use? (c) Jeff uses 61 tiles altogether to make a pattern in his sequence. What is the number of the pattern he makes?
- Owen has some tiles like these: He uses the tiles to make a series of patterns.
pattern number 1
pattern number 3
pattern number 2
pattern number 4 (a) Each new pattern has more tiles than the one before. The number of tiles goes up by the same amount each time. How many more tiles does Owen add each time he makes a new pattern? (b) How many tiles will Owen need altogether to make pattern number 6? (c) How many tiles will Owen need altogether to make pattern number 9? (d) Owen uses 40 tiles to make a pattern. What is the number of the pattern he makes? (KS3/98/Ma/Tier 4-6/P2)
10.2 Finding the Formula for a Linear Sequence
It is possible to determine a formula for linear sequences, i.e. sequences where the difference between successive terms is always the same. The first differences for the number pattern 11 14 17 20 23 26 ...
are 3 3 3 3 3 If we look at the sequence 3 n , i.e. the multiples of 3, and compare it with our original sequence our sequence 11 14 17 20 23 26 sequence 3 n 3 6 9 12 15 18 we can see easily that the formula that generates our number pattern is n th term of sequence = 3 n + 8 i.e. u (^) n = 3 n + 8 If, however, we had started with the sequence 38 41 44 47 50 53 ... the first differences would still have been 3 and the comparison of this sequence with the sequence 3 n our sequence 38 41 44 47 50 53 sequence 3 n 3 6 9 12 15 18 would have led to the formula u (^) n = 3 n + 35. In the same way, the sequence
- 7 – 4 – 1 2 5 8 ... also has first differences 3 and the comparison our sequence – 7 – 4 – 1 2 5 8 sequence 3 n 3 6 9 12 15 18 yields the formula u (^) n = 3 n − 10. From these examples, we can see that any sequence with constant first difference 3 has the formula u (^) n = 3 n + c where the adjustment constant c may be either positive or negative. This approach can be applied to any linear sequence, giving us the general rule that: If the first difference between successive terms is d, then u (^) n = d × n + c
The first term can be used to form an equation to determine c :
2 = 5 × 1 + c 2 = 5 + c c = – 3
So the formula will be,
u (^) n = 5 n − 3 Note that the constant term, c, is given by c = first term −first difference
Example 3
Determine a formula for the sequence: 28, 25, 22, 19, 16, 13, ...
Solution
First consider the differences between the terms,
28 25 22 19 16 13 ...
Here the difference is negative because the terms are becoming smaller. Using the difference as – 3 gives, u (^) n = − 3 n + c
The first term is 28, so
28 = − 3 × 1 + c 28 = − 3 + c c = 31
The general formula is then,
u (^) n = − 3 n + 31
or
u (^) n = 31 − 3 n
Exercises
- For the sequence, 7, 11, 15, 19, ... (a) calculate the difference between successive terms, (b) determine the formula that generates the sequence.
- Determine the formula for each of the following sequences: (a) 6, 10, 14, 18 , 22 , ... (b) 11, 13, 15, 17, 19, ... (c) 9, 16, 23, 30, 37, ... (d) 34, 56, 78, 100, 122, ... (e) 22, 31, 40, 49, 58, ...
- One number is missing from the following sequence: 1, 6, 11, , 21, 26, 31 (a) What is the missing number? (b) Calculate the difference between successive terms. (c) Determine the formula that generates the sequence.
- Determine the general formula for each of the following sequences: (a) 1, 4, 7, 10, 13, ... (b) 2, 6, 10, 14, 18, ... (c) 4, 13, 22, 31, 40, ... (d) 5, 15, 25, 35, 45, ... (e) 1, 20, 39, 58, 77, ...
- For the sequence, 18, 16, 14, 12, 10, ... (a) calculate the difference between successive terms, (b) determine the formula that generates the sequence.
- Determine the general formula for each of the following sequences: (a) 19, 16, 13, 10, 7, ... (b) 100, 96, 92, 88, 84, ... (c) 41, 34, 27, 20, 13, ... (d) 66, 50, 34, 18, 2, ... (e) 90, 81, 72, 63, 54, ...
(b) Copy and complete this table by writing expressions :
(c) Write an expression to show the total number of tiles in pattern number n. Simplify your expression. (d) A different series of patterns is made with tiles.
pattern number 1 pattern number (^2) numberpattern 3
The series of patterns continues by adding each time. For this series of patterns, write an expression to show the total number of tiles in pattern number n. Show your working and simplify your expression. (KS3/98/Ma/Tier 5-7/P1)
10.3 Second Differences and Quadratic Sequences In section 10.2 we dealt with sequences where the differences between the terms was a constant value. In this section we extend this idea to sequences where the differences are not constant.
Example 1
(a) Calculate the first 6 terms of the sequence defined by the quadratic formula, u (^) n = n^2 + n − 1 (b) Calculate the first differences between the terms. (c) Comment on the results you obtain.
pattern number expression for expression for the number of the number of grey tiles white tiles
n
Solution
(a) Substituting n = 1 gives,
u (^) 1 = 1 2 + 1 − 1 = 1 For n = 2 , u (^) 2 = 2 2 + 2 − 1 = 5 For n = 3 , u (^) 3 = 3 2 + 3 − 1 = 11 For n = 4 , u (^) 4 = 4 2 + 4 − 1 = 19 For n = 5 , u (^) 5 = 5 2 + 5 − 1 = 29 For n = 6 , u (^) 6 = 6 2 + 6 − 1 = 41 So the first 6 terms are, 1, 5, 11, 19, 29, 41
(b) The differences can now be calculated,
1 5 11 19 29 41
4 6 8 10 12
(c) Note that the differences between the first differences are constant. They are all equal to 2. These are called the second differences , as shown below. Sequence 1 5 11 19 29 41
First differenc es 4 6 8 10 12
Second differences 2 2 2 2
Example 3
Determine a formula for the general term of the sequence, 2, 9, 20, 35, 54, ...
Solution
Consider the first and second differences of the sequence:
2 9 20 35 54 ...
7 11 15 19
4 4 4 As the second differences are constant and equal to 4, the formula will begin u (^) n = 2 n^2 +...
To determine the rest of the formula, subtract 2 n^2 from each term of the sequence, as shown below:
Sequence 2 9 20 35 54 ... 2 n^22 8 18 32 New sequence 0 1 2 3 4
1 1 1 1
The new sequence has a constant difference of 1 and begins with 0, so for this sequence the formula is n − 1. Combining this with the 2 n^2 gives
u (^) n = 2 n^2 + n − 1
Example 4
(a) Calculate the first and second differences for the sequence, 4, 1, 0, 1, 4, 9, ... (b) Use the differences to determine the next 2 terms of the sequence. (c) Determine a formula for the general term of the sequence.
Solution
(a) 4 1 0 1 4 9 ...
2 2 2 2
(b) Extending the sequences above gives, 4 1 0 1 4 9 16 25 ...
2 2 2 2 2 2 (c) As the second differences are constant and all equal to 2, the formula will contain an ' n^2 ' term, and be of the form u (^) n = n^2 + a n + b We must now determine the values of a and b. The easiest way to do this is to subtract n^2 from each term of the sequence, to form a new, simpler sequence. our sequence 4 1 0 1 4 9 sequence n^2 1 4 9 16 25 new sequence 3 – 3 – 9 – 15 – 21 – 27 The new sequence 3 – 3 – 9 – 15 – 21 – 27 ...
- 6 – 6 – 6 – 6 – 6 has constant first differences of – 6 so will be given by − 6 n + b. Using the first term gives, 3 = − 6 × 1 + b b = 9
Thus the formula for the simpler sequence is − 6 n + 9. Now combining this with the n^2 term gives, u (^) n = n^2 − 6 n + 9
- Determine the formula for the general term of each of the following sequences: (a) 3, 17, 39, 69, 107, ... (b) 5, 18, 37, 62, 93, ... (c) 9, 23, 45, 75, 113, ... (d) – 4, 12, 38, 74, 120, ...
- (a) Calculate the second differences for the sequence, 9, 4, – 5, – 18, – 35, ... (b) Determine the formula for the general term of the sequence. (c) Hence show that the 20th term of the sequence is –770.
- Determine the formula for the general term of the sequence, 6, 10, 12, 12, 10, 6, ...
- (a) Calculate the first, second and third differences for the sequence, 6, 13, 32, 69, 130, 221, ... (b) Determine a formula for the general term of the sequence.
- This is a series of patterns with grey and black tiles.
pattern number 1
pattern number (^2) numberpattern 3 (a) How many grey tiles and black tiles will there be in pattern number 8? (b) How many grey tiles and black tiles will there be in pattern number 16? (c) How many grey tiles and black tiles will there be in pattern number P? (d) T = total number of grey tiles and black tiles in a pattern P = pattern number Use symbols to write down an equation connecting T and P. (Ks3/96/Ma/Tier 6-8/P2)
10.4 Special Sequences
Before going on to look at harder examples, we list some of the important sequences that you should know: 1, 3, 5, 7, 9, 11, 13, ... the odd numbers u (^) n = 2 n − 1 2, 4, 6, 8, 10, 12, 14, ... the even numbers u (^) n = 2 n 1, 4, 9, 15, 25, 36, 49, ... the square numbers u (^) n = n^2 1, 8, 27, 64, 125, 216, 343, ... the cube numbers u (^) n = n^3 1, 3, 6, 10, 15, 21, 28, ... the triangular numbers u (^) n = 12 n n ( + (^1) ) There is one other important sequence, namely the prime numbers 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, ... Note : there is no formula for calculating the n th prime number. We now look at other, harder sequences generated by algebraic rules.
Example 1
(a) Write down the next 3 terms of the sequence, 1, 1, 2, 3, 5, 8, 13, ... (b) Determine a formula for calculating the n th term.
Solution
(a) Use the first differences to extend the sequence: 1 1 2 3 5 8 13 ...
0 1 1 2 3 5 Note that the first differences, ignoring the first 0, are in fact the actual sequence itself. These can then be used to extend the sequence: 1 1 2 3 5 8 13 21 34 55 ...
0 1 1 2 3 5 8 13 21
