4 The first 3 terms in the expansion of a x x 1 3 3 5 + - ` b j l , in ascending powers of x, can be written in the form bx cx 27 2 + + , where a, b and c are integers. Find the values of a, b and c. [8]

4 ( ) 3 3 2 2 3 3 3   + = + + +   a x a a x ax x B1

4 (a) Find the first three non-zero terms in the expansion of x 2 4 2 6 - e o in ascending powers of x. Simplify each term. [3] (b) Hence find the term independent of x in the expansion of x x 2 4 3 1 2 6 2 2 - - e e o o . [3]

4(a) 2 4 64 48 15 − + x x 3 B1 for 64 B1 for 2 48 − x B1 for 4 15x 4(b) 2 4 6 1 9 − + x x B1 ( ) ( ) ( ) 64 9 48 6 15 × + − × − + their their their M1 For considering terms independent of x, must have 3 terms 879 A1

4 (a) Find the first three terms, in ascending powers of x2, in the expansion of x 2 1 3 2 2 8 - e o . Write your coefficients as rational numbers. [3] (b) Find the coefficient of x2 in the expansion of x x x 2 1 3 2 2 1 2 8 2 - + e b o l . [3]

4(a) 2 4 1 7 256 24 36 − + x x 3 B1 for 1 256 B1 for 2 24 −x B1 for 4 7 36 x 4(b) 2 2 1 4 4 + + x x soi B1 Coefficient of 2 x 1 4 256 1 4 24 7 36 their their their their their   ×       + × −       +    M1 Allow one sign error, but must have 3 terms in 2 x only, with an attempt at addition. 25 576 A1

5 A geometric progression is such that its sum to 4 terms is 17 times its sum to 2 terms. It is given that the common ratio of this geometric progression is positive and not equal to 1. (a) Find the common ratio of this geometric progression. [3] (b) Given that the 6th term of the geometric progression is 64, find the first term. [2] (c) Explain why this geometric progression does not have a sum to infinity. [1]

5(a) ( ) ( ) 4 2 1 1 17 1 1 − − = − − a r a r r r M1 Allow equivalents Allow if ‘a’ terms missing (assume to have been cancelled) ( )( ) ( ) 2 2 2 1 1 17 1 − + = − r r r or better 4 2 17 16 0 − + = r r oe 3 2 16 16 0 + − − = r r r oe M1 Dep M1 for a correct simplified equation in r only r = 4 only, from correct working A1 5(b) 5 64 = ar M1 For use of 5 a r with their positive r 0.0625 = a or 1 16 A1 Must be exact A0 if 4 = r not from correct working in (a) 5(c) Because 1 > r oe B1 FT on their 1 > r Must have a value for r

6 (a) A geometric progression has first term 10 and sum to infinity 6. (i) Find the common ratio of this progression. [2] (ii) Hence find the sum of the first 7 terms, giving your answer correct to 2 decimal places. [2] (b) The first three terms of an arithmetic progression are log 3 x , ( ) log 3 x 2 , ( ) log 3 x 3 . (i) Find the common difference of this progression. [1] (ii) Find, in terms of n and log 3 x , the sum to n terms of this progression. Simplify your answer. [2] (iii) Given that the sum to n terms is log 3081 3 x , find the value of n. [2] (iv) Hence, given that the sum to n terms is also equal to 1027, find the value of x. [2]

6(a)(i) 10, 6 1 = = − a a r 10 6 6 = −r M1 For use of first term and sum to infinity to obtain an equation in r only 2 3 = − r A1 6(a)(ii) ( ) ( ) 7 7 1 10 1 − = − their r S their r M1 For sum formula with . 1 < their r 7 6.35 = S A1 6(b)(i) log 3 x B1 6(b)(ii) ( ) ( ) 2log 3 1 log 3 2 = + − n x x n S n M1 For use of sum formula with their (i) ( ) 1 1 1 log 3, log 3 , log 3 2 2 2 + + + n n x x x n n n n A1 Allow other similar equivalents 6(b)(iii) ( )1 3081 2 + = n n M1 For a correct attempt to solve their (ii) 3081log 3 = x to obtain an answer for n. Must be a 3 term quadratic in n only. 78 = n A1 6(b)(iv) ( ) 78 1027 79 log 3 2 = x or 3081 log 3 x M1 For using their 78 in a sum equation or using 3081 to obtain x 27 = x A1