What you need to know To recognise GP’s and use nth term and sum of n terms formulae to solve problems To know about the sum of an infinite GP where How to use the binomial expansion of

Geometric Sequences eg. 10 20 40 80 81 27 9 3 x 2 x 2 x 2 x ⅓ x ⅓ x ⅓ 81 27 9 3 x 2 x 2 x 2 x ⅓ x ⅓ x ⅓ first term = 10 first term = 81 common ratio = 2 common ratio = ⅓

Geometric kth term = a r k - 1 Term S n = a ( 1 – r n ) ( 1 – r ) Sum of terms S n = a ( 1 – r n ) ( 1 – r ) S  = a 1 - r Sum to infinity

Find the common ratio, the first term and the 10th term Example 1 Given that the 2nd term of a positive geometric sequence is 4 and the 4th term is 8. Find the common ratio, the first term and the 10th term 2nd term = 4  ar = 4 Ignore the negative root 4th term = 8  ar 3 = 8 dividing gives r 2 = 2  common ratio = √2 substituting into first equation gives a √2 = 4 a = 4 √2 a = 4√2 2  First term = 2√2 10th term = ar 9 = 2√2 ( √ 2 ) 9 = 2 ( √ 2 ) 10 = 2 (2 ) 5 = 2 6 10th term = 64

Example 2 How many terms are there in the geometric sequence 0 nth term = a r n - 1  There are 10 terms

Example 3 The numbers 3, x and ( x + 6 ) form the first three terms of a positive geometric sequence. Find the possible values of x and the 10th term of the sequence. x = x + 6 3 x 10th term = a r 9 x 2 = 3 ( x + 6 ) = 3 x 2 9 x 2 = 3 x + 18  10th term is 1536 x 2 - 3 x - 18 = 0 ( x – 6 ) ( x + 3 ) = 0 x = 6 or -3 x = 6 ( since there are no negative terms )

Infinite Geometric Series Consider the series S = 3 + 1.5 + 0.75 + 0.375 + 0.1875 + ….. Summing the terms one by one gives 3, 4.5, 5.25, 5.625, 5.8125, … No matter how many terms you take, the sum never exceeds a certain number. We call this number the limit of the sum. We say the series is convergent. S  = a 1 - r ( The limit only exists if -1 < r < 1 ) So for above example S  = 3 1 – 0.5 = 6

Example 4 The first term of a GP is 10 and the common ratio is 0. 8 Example 4 The first term of a GP is 10 and the common ratio is 0.8. (a) Find the 4th term and the sum of the first 20 terms

(b) The sum of the first N terms is SN and the sum to infinity is S (b) The sum of the first N terms is SN and the sum to infinity is S. Show the inequality S – SN < 0.01 can be written as 0.8N < 0.0002 and use logarithms to find the smallest possible value of N Smallest N is 39

This button is on your calculator Pascal’s triangle 1 1 1 2 1 1 3 3 1 1 4 6 4 1 Binomial expansion (a+b)n can be expanded by using the numbers from Pascal’s triangle, decreasing powers of a and increasing powers of b This button is on your calculator

Example 6 Expand (1+2x)4 in ascending powers of x   (1+2x)4 =1 + 8x + 24x2 + 32x3 +16x4 1 4 6 14 13 12 11 10 (2x)0 (2x)1 (2x)2 (2x)3 (2x)4 Put the powers of the first term descending Put the numbers from Pascal’s triangle in first Put the powers of the other term ascending

Put the numbers from Pascal’s are on the calculator Example 7 Find the coefficient of t4 in the expansion of (3 – 2t)8 1 8 38 37 34 (-2t)0 (-2t)1 (-2t)4 Put the numbers from Pascal’s are on the calculator t4 term is …90720t4 …..Coefficient of t4 is 90720

Example 8 Expand (1-x)5 as far as the term in x2. Hence find an approximation to 0.95 1 5 10 15 14 13 (-x)0 (-x)1 (-x)2

Example 9 Find the Binomial expansion of (2x + 5)4 , simplifying the terms 1 4 6 (2x)4 (2x)3 (2x)2 (2x)1 (2x)0 50 51 52 53 54

Example 9 Find the Binomial expansion of (2x + 5)4 , simplifying the terms Hence show that (2x + 5)4 – (2x – 5)4 can be written as 320x3 + kx where the value of the constant k is to be stated. k =2000

Verify that x = 2 is a root of the equation (2x + 5)4 – (2x – 5)4 = 3680x – 800 and find the other possible values of x.

x 4x3 -2

4x2 8x -5 x 4x3 8x2 -5x -2 -8x2 -16x 10 Now factorise the quadratic factor

Summary To recognise GP’s and use nth term and sum of n terms formulae to solve problems To know about the sum of an infinite GP where How to use the binomial expansion of