GCE Mathematics, A2 Unit 3 Pure Mathematics B
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GCE Mathematics, A2 Unit 3 Pure Mathematics B Topic 3.4 - Sequences and series The binomial theorem - recap from AS Unit 1 The binomial expansion of (a + b) n , for n a positive integer, is given by (a + b) n = a n + n 1 ( ) a n–1 b + ( n 2 ) a n–2 b 2 + … + ( n n -1 ) ab n–1 + b n This is given in the Formula Booklet. Note that as the powers of a decrease, the powers of b increase, and there will be n + 1 terms in the expansion. n r ( ) = n C r = n! r!(n – r)! n! means n factorial remember 0! = 1 Examples 5! = 5 × 4 × 3 × 2 × 1 = 120 6 C 4 = 6! 4!2! = 15 7 3 ( ) = 7! 3!4! = 35 Pascal’s triangle can also be used to determine the coefficients in a binomial expansion. Binomial expansion for negative and fractional values of n The binomial expansion of (a + bx) n , where n is negative or a fraction, is given by (a + bx) n = a n + n 1 ( ) a n–1 bx + ( n 2 ) a n–2 (bx) 2 + … which is only valid for | bx a | < 1. Replacing a and b by 1, gives (1 + x) n = 1 + nx + n(n – 1) 2! x 2 + n(n – 1)(n – 2) 3! x 3 + … valid for |x| < 1. This is given in the Formula Booklet. (1 + x) –6 = 1 – 6x – 6(-7) 2! x 2 – 6(-7)(-8) 3! x 3 + … = 1 – 6x + 21x 2 – 56x 3 + … valid for |x| < 1 (1 – 4x) –3 = 1 – 3(–4x) – 3(-4) 2! (–4x) 2 – 3(-7)(-5) 3! (–4x) 3 + … = 1 + 12x + 96x 2 + 640x 3 + … valid for |4x| < 1 Arithmetic sequences In an arithmetic sequence, consecutive terms differ by a fixed amount known as the common difference. The terms can be written as t 1 = a, t 2 = a + d, t 3 = a + 2d, etc, where a is the first term, and d is the common difference. The formula for the nth term is given by t n = a + (n – 1)d. The proof of the sum of an arithmetic sequence (S n ) could be asked for in an examination. S n = a + (a + d ) + (a + 2d ) + … + (a + (n – 1)d ) S n = (a + (n – 1)d ) + (a + (n – 2)d ) + … + (a + d) + d 2S n = (2a + (n – 1)d ) + (2a + (n – 1)d ) + … + (2a + (n – 1)d ) 2S n = n(2a + (n – 1)d ) S n = n 2 (2a + (n – 1)d ) The summation sign ⅀ can also be used. Geometric sequences In a geometric sequence, the ratio of consecutive terms is a fixed number known as the common ratio. The terms can be written as t 1 = a, t 2 = ar, t 3 = ar 2 etc, where a is the first term, and r is the common ratio. The formula for the nth term is given by t n = ar n–1 . The proof of the sum of a geometric sequence (S n ) could be asked for in an examination. S n = a + ar + ar 2 + ar 3 + … + ar n–1 rS n = ar + ar 2 + ar 3 + ar 4 +… + ar n rS n - S n = ar n – a S n = a (r n – 1) (r – 1) or S n = a (1 – r n ) (1 – r) , r ≠ 1 When |r| < 1, r n → 0 as n → ∞, giving S ∞ = a 1 – r . Periodic sequences A periodic sequence is generated by repeating a set of numbers. For example, 1, 2, 3, 4, 1, 2, 3, 4, 1, … Recurrence sequences x n+1 = f(x n ) A recurrence sequence is one in which successive terms are generated from a function involving the preceding terms. Increasing/decreasing sequences A sequence is increasing if each term is greater than the preceding term, and it is decreasing if each term is less than the preceding term.
Transcript of GCE Mathematics, A2 Unit 3 Pure Mathematics B
GCE Mathematics, A2 Unit 3 Pure Mathematics BTopic 3.4 - Sequences and series
The binomial theorem - recap from AS Unit 1
The binomial expansion of (a + b)n, for n a positive integer, is given by
(a + b)n = an + n1( )an–1b + (n
2)an–2b2 + … + ( nn -1 )abn–1 + bn
This is given in the Formula Booklet. Note that as the powers of a decrease, the powers of b increase, and there will be n + 1 terms in the expansion.
nr( ) = nCr = n!
r!(n – r)!
n! means n factorial remember 0! = 1
Examples
5! = 5 × 4 × 3 × 2 × 1 = 120 6C4 = 6!
4!2! = 1573( )= 7!
3!4! = 35
Pascal’s triangle can also be used to determine the coefficients in a binomial expansion.
Binomial expansion for negative and fractional values of n
The binomial expansion of (a + bx)n, where n is negative or a fraction, is given by
(a + bx)n = an + n1( )an–1bx + (n
2)an–2(bx)2 + …
which is only valid for |bxa | < 1.
Replacing a and b by 1, gives
(1 + x)n = 1 + nx + n(n – 1)2! x2 + n(n – 1)(n – 2)
3! x3 + …
valid for |x| < 1. This is given in the Formula Booklet.
(1 + x)–6 = 1 – 6x – 6(-7)2! x2 – 6(-7)(-8)
3! x3 + …
= 1 – 6x + 21x2 – 56x3 + … valid for |x| < 1
(1 – 4x)–3 = 1 – 3(–4x) – 3(-4)2! (–4x)2 – 3(-7)(-5)
3! (–4x)3 + …
= 1 + 12x + 96x2 + 640x3 + … valid for |4x| < 1
Arithmetic sequences
In an arithmetic sequence, consecutive terms differ by a fixed amount known as the common difference.
The terms can be written as t1 = a, t2 = a + d, t3 = a + 2d, etc, where a is the first term, and d is the common difference.
The formula for the nth term is given by tn = a + (n – 1)d.
The proof of the sum of an arithmetic sequence (Sn) could be asked for in an examination.
Sn = a + (a + d ) + (a + 2d ) + … + (a + (n – 1)d ) Sn = (a + (n – 1)d ) + (a + (n – 2)d ) + … + (a + d) + d
2Sn = (2a + (n – 1)d ) + (2a + (n – 1)d ) + … + (2a + (n – 1)d ) 2Sn = n(2a + (n – 1)d )
Sn = n2(2a + (n – 1)d )
The summation sign ⅀ can also be used.
Geometric sequences
In a geometric sequence, the ratio of consecutive terms is a fixed number known as the common ratio.
The terms can be written as t1 = a, t2 = ar, t3 = ar2 etc, where a is the first term, and r is the common ratio.
The formula for the nth term is given by tn = ar n–1.
The proof of the sum of a geometric sequence (Sn) could be asked for in an examination.
Sn = a + ar + ar2 + ar3 + … + arn–1
rSn = ar + ar2 + ar3 + ar4 +… + arn
rSn - Sn = arn – a
Sn = a (rn – 1)(r – 1) or Sn = a(1 – rn)
(1 – r) , r ≠ 1
When |r| < 1, rn → 0 as n → ∞, giving S∞ = a1 – r.
Periodic sequences
A periodic sequence is generated by repeating a set of numbers. For example, 1, 2, 3, 4, 1, 2, 3, 4, 1, …
Recurrence sequences xn+1 = f(xn)
A recurrence sequence is one in which successive terms are generated from a function involving the preceding terms.
Increasing/decreasing sequences
A sequence is increasing if each term is greater than the preceding term, and it is decreasing if each term is less than the preceding term.
