Use the binomial theorem for expansion of — Paper 2 · IGCSE Mathematics - Additional 0606

6 The first three terms of the binomial expansion of 2 - ax n are 64 - 16 bx + 100 bx 2 . Find the value of each of the integers n, a and b. [7]

7 marks

Mark scheme: 6 64 = 2 n M1 n = 6 A1 their (6 1) M1 their 6 ( 2 ) −× ( − a ) = − 16b oe 6 × (6 − 1) their (6 − 2) 2 M1 their ( 2 ) × ( − a ) = 100b oe 2 attempts to solve DM1 dep on both M1 marks being awarded; must have correctly or correct FT eliminated one unknown a = 5 A1 b = 60 A1

This question in 0606/23 May/June 2017

8 (a) (i) Given that x 2 - = x 16 - 4x 13 + qx 10 + rx 7 + f, find the value of each of the constants p, px q and r. [3] 8 2 1 (ii) Explain why there is no term independent of x in the binomial expansion of x - . [1] e px o n x (b) In the binomial expansion of 1 - , where n is a positive integer, the coefficient of x is 30. Form e 2 o an equation in n and hence find the value of n. [4]

8 marks

Mark scheme: 8(a)(i) p = 2, q = 7, r = −7 B3 B1 for each or M1 for any two of 2 6  1  2 7  1  8 × 7 2 x x 8 −  , ( )  −  , ( )   px  2  px  3 8 × 7 × 6 2 x −  ( 5  1  )  3 × 2  px  or better 8(a)(ii) Valid explanation B1 8(b) 2 B1 n ( n − 1)  1   −  [ x ] seen or implied 2  2   n ( n − 1)  1  2  M1 their   −   = 30  2  2     n 2 − n − 240 = 0 A1 16 A1

This question in 0606/23 May/June 2019

10 (i) Expand (3 + )x 4 evaluating each coefficient. [3] p 4 In the expansion of x - (3 + x) the coefficient of x is zero. b x l (ii) Find the value of the constant p. [2] (iii) Hence find the term independent of x. [1] (iv) Show that the coefficient of x2 is 90. [2]

8 marks

Mark scheme: 10(i) 81 + 108 x + 54 x 2 + 12 x 3 + x 4 B3 B1 for coefficients B1 for powers B1 for all Correct 10(ii) Identify and select two terms in x and M1 81 – 54p = 0 equate to zero p = 1.5 A1 10(iii) Constant term = –108p = –162 A1 FT using their p 10(iv) Correctly identify two terms in x2 M1 x2 term = 108 – 12p 108 – 18 = 90 A1

This question in 0606/21 Oct/Nov 2019

3 The first four terms in the expansion of (1 + ax) 5 (2 + bx) are 2 + 32x + 210x 2 + cx 3 , where a, b and c are integers. Show that 3a 2 - 16a + 21 = 0 and hence find the values of a, b and c. [8]

8 marks

Mark scheme: 3 (1 + ax)5 = 1+ 5ax + 10a2x2 + 10a3x3 soi B1 4 terms not nCr notation [2] + (10a + b)x + (5ab + 20a2)x2 M1 obtain expansion with 2 terms in x, 2 terms in x2 equate terms in x and x2 to give two M1 equations in a and b each consisting of three terms 10a + b = 32 A1 correct equations imply previous two 5ab + 20a2 = 210 M marks eliminate b M1 obtain 3a2 – 16a + 21 = 0 correctly A1 answer given a = 3 and b = 2 B1 c = 720 only B1 no additional answers

This question in 0606/23 Oct/Nov 2019

8 (a) Expand ( 2 - x) 5 , simplifying each coefficient. [3] e ( 2 - )x 5 # e 80 x -x 5 (b) Hence solve 4 = e . [4] e 10x + 32

7 marks

Mark scheme: 8(a) 32 – 80x + 80x2 – 40x3 + 10x4 – x5 B3 B2 for any four or five terms correct or B1 for any three terms correct or M1 for a fully correct but unsimplified expansion 8(b) Combines powers sufficiently to be M1 able to take logs or applies correct log laws For making use of their expansion M1 from part (a) 40x2 (2 – x) [= 0] oe M1 FT their (a) if possible x = 0, x = 2 cao A1

This question in 0606/21 May/June 2020

9 DO NOT USE A CALCULATOR IN THIS QUESTION. 1 6 (a) Find the term independent of x in the binomial expansion of 3x - . [2] b x l x n (b) In the expansion of 1 + the coefficient of x4 is half the coefficient of x6. Find the value of b 2 l the positive constant n. [6]

8 marks

Mark scheme: 9(a) –540 B2 3 6 × 5 × 4 3  1  B1 for ( 3 x )  −  oe 3!  x  9(b) 6 B1 n ( n − 1)( n − 2)( n − 3)( n − 4)( n − 5)  1  ×  6!  2  4 B1 n ( n − 1)( n − 2)( n − 3)  1  ×  4!  2  Forms a correct equation with their M1 coefficients in terms of n Simplifies their equation to M1 (n – 4)(n – 5) = 240 or better Factorises or attempts to solve their 3-term M1 quadratic n = 20 A1

This question in 0606/23 May/June 2020

5 The first three terms in the expansion of a + bx ( 1 + x) are 32 - 208x + cx 2 . Find the value of each of the integers a, b and c. [7]

7 marks

Mark scheme: 5 a 5 + 5 a 4 bx + 10 a 3 b 2 x 2 2 B1 for powers or for coefficients 5 5 4 3 2 4 2 2 M1 for multiplying to obtain 5 terms a + a + 5a b x + 10 a b + 5a b x ( ) ( ) A1 for all correct a 5 = 32 → a = 2 A1 32 + 80b = −208 → b = −3 A1 10 × 8 × 9 + 5 × 16 ×−=3 c → c = 480 A1

This question in 0606/21 Oct/Nov 2020

9 (a) In the expansion of 2 k - , where k is a constant, the coefficient of x2 is 160. Find the value k of k. [3] 6 (b) (i) Find, in ascending powers of x, the first 3 terms in the expansion of 1 + 3x , simplifying ` j the coefficient of each term. [2] 6 2 (ii) When 1 + 3x a + x is written in ascending powers of x, the first three terms are ` j ` j 4 + 68x + bx 2 , where a and b are constants. Find the value of a and of b. [3]

8 marks

Mark scheme: 9(a) Identifies the correct term: B1 5 3  1  2 2 [× x ] oe, soi C 2 × (2 k ) ×  −   k  8 k 3 M1 FT only for correct term with bracketing 10 × = 160 soi errors; condone one slip in simplification 2 k k = 2 nfww A1 9(b)(i) 1 + 18x + 135x2 B2 B1 for any 2 terms correct or for all 3 correct terms listed but not summed or M1 for a correct unsimplified expansion e.g. : 1 + 6(3x) + 15(3x)2 9(b)(ii) Uses constant/coefficient of x to B2 B1 for both a = 2 and −2 or find a = −2 only 17 for both a = and −2 9 b = 469 only B1 FT their calculated value of a

This question in 0606/22 Feb/March 2021

1 Using the binomial theorem, expand ( 1 + e 2 x) 4 , simplifying each term. [2]

2 marks

Mark scheme: Question Answer Marks Partial Marks 1 1 + 4e 2 x + 6e 4 x + 4e 6 x + e 8 x B2 mark final answer for B2 B1 for any 3 correct simplified terms in a sum or all 5 simplified terms listed but not summed or for a correct, simplified expansion that is not their final answer or M1 for a correct unsimplified expansion e.g. 2 x 4 × 3 2 x 2 4 × 3 × 2 2 x 3 2 x 4 1 + 4e + (e ) + (e ) + (e ) 2 6 If 0 scored, SC1 for a complete, correct, simplified expansion as final answer found by multiplying out the brackets

This question in 0606/22 May/June 2021

2 (a) Expand ( 2 - 3)x 4 , evaluating all of the coefficients. [4] 4 a (b) The sum of the first three terms in ascending powers of x in the expansion of ( 2 - 3)x 1 + b x l 32 is + b + cx , where a, b and c are integers. Find the values of each of a, b and c. [4] x

8 marks

Mark scheme: 2(a) 16 − 96 x + 216 x 2 − 216 x 3 + 81x 4 B4 Mark final answer for B4 B3 for any 4 correct simplified terms in a sum or for all 5 simplified terms listed but not summed or for a correct simplified expansion that is not their final answer or B2 for any 3 correct simplified terms in a sum or for 4 correct simplified terms listed but not summed or B1 for any 2 correct simplified terms in a sum or for 3 correct simplified terms listed but not summed or M1 for correct unsimplified expansion 2 4 + 4 × 23 ( −3 x ) + 6 × 2 2 ( −3 x ) 2 3 4 +4 × 2 ( −3 x ) + ( −3 x ) 2(b) 2  a  B1 their 16 − 96 x + 216 x ….. ×  1 +  ( )  x  FT Expansion using their (a) a = 16 − 96 x + 16 − 96 a + 216 ax … soi x a = 2 B1 a FT their 16 x b = − 176 B1 c = 336 B1

This question in 0606/22 Oct/Nov 2021

6 (a) (i) Use the binomial theorem to expand ( 1 + 3)x 7 in ascending powers of x, as far as the term in x3. Simplify each term. [2] (ii) Show that your expansion from part (i) gives the value of .1037 as 1.23 to 2 decimal places. [2] 15 x 4 2 (b) Find the term independent of x in the expansion of + . [2] e 2 x o

6 marks

Mark scheme: 6(a)(i) 1 + 21x + 189 x 2 + 945 x 3 B2 B1 for three out of the four terms correct or for a correct answer seen then spoilt If 0 scored then SC1 for 1, 21x ,189 x 2 , 945 x 3 seen but not summed 6(a)(ii) 1 + 21(0.01) + 189(0.01) 2 + 945(0.01) 3 B2 M1 for use of x = 0.01 oe in their expansion seen or implied by or 1 + 0.21 + 0.0189 + 0.000945 oe e.g. 1 + 21(0.01) + 189(0.01) 2 + 945(0.01) 3 leading to 1.229[845 = 1.23] cao or 1 + 0.21 + 0.0189 + 0.000945 OR 1.229845 without working or from working that is not fully correct 6(b) 4 3 12 M1 15  x   2  C12 ×   ×   oe  2   x  232960 A1

This question in 0606/22 Feb/March 2022

5 (a) (i) Write down, in ascending powers of x, the first three terms in the expansion of ( 1 + 4x) n . Simplify each term. [2] (ii) In the expansion of ( 1 + 4x) n ( 1 - 4x) the coefficient of x2 is 6032. Given that n 2 0, find the value of n. [3] x 8 10 (b) Find the term independent of x in the expansion of - 4 . [2] e 2 x o

7 marks

Mark scheme: 5(a)(i) 1 + 4nx + 8n(n – 1)x2 B2 B1 for any two correct terms or all 3 correct but listed not summed 5(a)(ii) 8n(n – 1) – 16n M1 FT from (i) identifying correct terms and combining their coefficient of x2 – 4  their coefficient of x Solves or factorises their 3-term quadratic in n only M1 Forms a 3-term quadratic = 0 and solves except allow ‘= a constant’ if they go on to complete to square n = 29 only A1 5(b) 8 2 M1 Must be clearly identified not 10  x   8  C 2     4  soi in expansion  2   x  45 A1 11.25 or 4

This question in 0606/21 May/June 2022

6 The first four terms in ascending powers of x in the expansion ( 3 + ax) 4 can be written as 2 3 3 81 + bx + cx + x . Find the values of the constants a, b and c. [6] 2

6 marks

Mark scheme: 6 81 + 108ax + 54a 2 x 2 + 12a 3 x 3 soi M3 M2 for any 3 correct terms or 2 correct equations or M1 for any 2 correct terms, 1 3 3 2 or 12 a = b = 108a c = 54 a soi correct equation or for correct but 2 insufficiently simplified expansion e.g. 4 3 4  3 2 2 3 + 4  3  ax +  3  ( ax ) 2 4 3 2 3 + 3 ( ax ) 3  2 1 A1 a = oe 2 b = 54 A1 FT 108  their a, providing at least M1 awarded 27 A1 FT 54  (their a)2, providing at least c = oe 2 M1 awarded

This question in 0606/23 Oct/Nov 2022

2 DO NOT USE A CALCULATOR IN THIS QUESTION. x 11 2 Expand and simplify , giving your answer with a rational denominator. [4] e 2 3 - 1 o

4 marks

Mark scheme: 2 11x 2 B1 12 + 1 − 4 3 2 M1 FT their expression of 11x 13 + 4 3 ( ) equivalent difficulty 13 − 4 3 13 + 4 3 ( )( ) A1 11x 2 13 + 4 3 2 2 ( ) 143 x + 44 3 x or 169 − 48 169 − 48 A1 mark final answer x 2 13 + 4 3 2 2 ( ) 13 x + 4 3 x or 11 11 Alternative method 2 (M1)  x 11(2 3 + 1)       (2 3 − 1)(2 3 + 1)  2 2 (A1)  x 11(2 3 + 1)   2 33 x + x 11    or       12 − 1 12 − 1     11x 2 (12 + 1 + 4 3) 132 x 2 + 11x 2 + 4 363 x 2 (A1) or 121 121 2 (A1) mark final answer x 13 + 4 3 ( ) 11

This question in 0606/22 Feb/March 2023

6 (a) (i) Find the first three terms in the expansion of 1 + , in ascending powers of x. Simplify the 7 coefficient of each term. [2] 5 n x (ii) The expansion of 7 ( 1 + )x 1 + , where n is a positive integer, is written in ascending b 7 l powers of x. The first two terms in the expansion are 7 + 89x . Find the value of n. [2] (b) In the expansion of ( k - 2 x) 8 , where k is a constant, the coefficient of x4 divided by the coefficient of x2 is 5. The coefficient of x is positive. Form an equation and hence find the value of k. [5] 8

9 marks

Mark scheme: 6(a)(i) 5 10 2 B2 B1 for any two correct terms or for the 1  x  x three terms listed but not summed 7 49 6(a)(ii) 12 B2  5  B1 for 7n + 5 = 89 or 7 n   89 oe    7  or nC1  12 6(b) 8C 4  k 4 ( 2) 4 [ x 4 ] oe or 1120k4 [x4] soi B1 8C 2  k 6 ( 2) 2 [ x 2 ] oe or 112k6 [x2] soi B1 1120 k 4 5 70  16  k 4 5 M1 FT providing at least B1 awarded and 6  or 6  oe, soi correct terms attempted 112 k 8 28 4 k 8 k 2 = 16 soi A1 [For coefficient of x to be positive k < 0, A1 therefore] k =  4

This question in 0606/22 May/June 2023

4 In this question a and b are integers. Three terms in the expansion of ( 2 + ax) 5 ( 1 + bx) are 32 + 112x - 240 x 2 . Find the values of a and b. [7]

7 marks

Mark scheme: 4 ( 2 + ax )5 = 25 + 5  2 4 ax + 10  23 a 2 x 2 + ... B1 ( 2 + ax )5 (1 + bx ) = M1 32 + 80ax + 80a2x2 + 32bx + 80abx2... 80a + 32b = 112 oe, isw A1 80 a 2 + 80 ab = −240 oe, isw A1 3a 2 − 7 a − 6 = 0 oe M1 (3a + 2)(a – 3) = 0 M1 a = 3 and b = –4 and no other values A1

This question in 0606/21 Oct/Nov 2023

10 The expansion of a + in ascending powers of x begins b 4 + 48b 3 x , where n, a and b are a positive integers. n 2 2 - 4 48 (a) Show that a = . [4] b n l (b) Given also that the third term is 1056 b 2 x 2, find the values of n, a and b. [6] Question 11 is printed on the next page.

10 marks

Mark scheme: 10(a) n 4 n −1 1 3 M1 a = b and na  = 48b oe  a  Eliminates b from one equation using the M1 dep previous M1 other equation e.g. a n − 2 48 = 3 n n a 4 Simplifies a terms e.g. A1 n 3 n 3 2 48 4 − 6  48  4 a −= or a =   n  n  Uses an appropriate power and completes to A1 the given form e.g. 2 2 3 2 3 n n 3 − 6 − 2 3  48   48  4 4 ) 2 ) = or ( a ( a =      n   n  n 2 − 4  48  2 → a =    n  10(b) Correct equation in a, b, n M1 n ( n − 1) n − 2 1 2  a  = 1056b oe, soi 2 a 2 Correct equation in a, n A1 n ( n − 1) n − 2 1 n2  a  = 1056 a oe 2 2 a − 4  A1  n( n − 1) n2   a = 1056 oe →  2  Correct equation in n only n ( n − 1)  48  2    = 1056 oe 2  n  n2 – 12n = 0 or n – 12 = 0 oe A1 n = 12 only A1 a = 4 only and b = 64 only A1

This question in 0606/22 Feb/March 2024

4 (a) Find and simplify the term independent of x in the expansion of x 2 - 3 . [2] 2x (b) DO NOT USE A CALCULATOR IN THIS PART OF THE QUESTION. 4 4 (i) Use the binomial theorem to show that 1 + 2 2 - 1 - 2 2 = k 2 , where k is an integer ` j ` j to be found. [4] 4 4 1 + 2 2 - 1 - 2 2 (ii) Hence write ` j ` j in the form a + b 2 , where a and b are integers. [2] 1 + 2

8 marks

Mark scheme: 4 2 4(a) 105 1   isw or 13.125 oe oe B1 for 10C 4 ( x 2 ) 6    3 8  2 x  4(b)(i) 1  4(2 2)  6(2 2) 2  4(2 2) 3  (2 2) 4 M1 soi or 1  4(  2 2)  6(  2 2) 2  4(  2 2) 3 ( 2 2) 4 soi 1  8 2  48  64 2  64 or A1 1  8 2  48  64 2  64 Correct difference stated or clearly M1 dep on sight of correct expansions with implied numerical coefficients 1  8 2  48  64 2  64  (1  8 2  48  64 2  64) 144 2 nfww A1 4(b)(ii)  their k 2 1  2 2 STRICT FT of their integer value of k  = 1  2 1  2  their k 2  2their k oe 1 B1 STRICT FT of their integer value of k simplified to 2(their k) – (their k) 2 their k 2 1  2 mark final answer for  1  2 1  2

This question in 0606/23 May/June 2024