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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.

GCE Mathematics, A2 Unit 3 Pure Mathematics B

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Page 1: 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.