Q3 Cambridge International Examinations

Cambridge International Examinations Cambridge International Advanced Subsidiary and Advanced Level May-June 2015 MATHEMATICS (US) Paper 1 Pure Mathematics 1 (P1) shahbaz ahmed June 2024 Relevant formulas and skills involved in the solution of the problem (a + b) = a + b (a + b)2 = a2 + 2ab + b2 (a + b)3 = a3 + 3a2 b + 3ab2 + b3 (a + b)4 = a4 + 4a3 b + 6a2 b2 + 4ab3 + b4 Using the formula Cr = n! r!(n−r)! C0 = 1! 0!(1−0)! n 1 = 1! 0!×1! By definition 1 n! = n(n − 1)(n − 2)....3 × 2 × 1 0! = 1 1! = 1 1 C0 = 1 1 =1 Similarly 1! 1!×0! 1 1 1 C1 = 1! 1!(1−1)! 2 C0 = 2! (2−0)!×0! = 2 2×1 =1 2 C1 = 2! (2−1)!×1! = 2 1×1 =2 2 C2 = 2! (2−2)!×2! = 2 0!×2 = 3 C0 = 3! (3−0)!×0! = 6 6×1 =1 3 C1 = 3! (3−1)!×1! = 6 2×1 = 6 2 =3 3 C2 = 3! (3−2)!×2! = 6 1×2 = 6 2 =3 3 C3 = 3! (3−3)!×3! = 6 1×6 = 6 6 =1 4 C0 = 4! (4−0)!×0! = 24 24×1 4 C1 = 4! (4−1)!×1! = 24 6×1 =4 4 C2 = 4! (4−2)!×2! = 24 2×2 = 24 4 =6 4 C3 = 4! (4−3)!×3! = 24 1×6 = 24 6 =4 4 C4 = 4! (4−4)!×4! = 24 0!×24 = = =1 2 1×2 = = 24 24 =1 =1 24 1×24 = 24 24 =1 Hence (a + b) =1 C0 a +1 C1 b (a + b)2 =2 C0 a2 +2 C1 ab +2 C2 b2 (a + b)3 =3 C0 a3 +3 C1 a2 b +3 C2 ab2 +3 C3 b3 (a + b)4 =4 C0 a4 +4 C1 a3 b +4 C2 a2 b2 +4 C3 ab3 +4 C4 b4 In general (a+b)n =n C0 an +n C1 an−1 b+n C2 an−2 b2 +...........+n Cr an−r br +........n Cn bn = T1 + T2 + T3 + ........ + Tr+1 + ........ + Tn =⇒ 2 Tr+1 =n Cr an−r br Q3 (i) Find the first three terms, in ascending powers of x, in the expansion of (a) (1 − x)6 , (b)(1 + 2x)6 (ii) Hence find the coefficient of x2 in the expansion of [(1 − x)(1 + 2x)]6 Solution (i)(a) (a + b)n = (1 − x)6 =⇒ a = 1, b = (−x), n = 6 Putting in the formula Tr+1 =n Cr an−r br Tr+1 =6 Cr 16−r (−x)r Tr+1 =6 Cr (−x)r Putting r = 0, 1, 2 T0+1 = T1 =6 C0 (−x)0 = 1 T1+1 = T2 =6 C1 (−x)1 = −6x T2+1 = T3 =6 C2 (−x)2 = 15x2 First three terms are: T1 + T2 + T3 = 1 − 6x + 15x2 (i)(b) (a + b)n = (1 + 2x)6 =⇒ a = 1, b = (2x), n = 6 Putting in the formula Tr+1 =n Cr an−r br Tr+1 =6 Cr 1n−r (2x)r Tr+1 =6 Cr (2x)r Putting r = 0, 1, 2 T0+1 = T1 =6 C0 (2x)0 = 1 3 T1+1 = T2 =6 C1 (2x)1 = 12x T2+1 = T3 =6 C2 (2x)2 = 15(4x2 ) = 60x2 First three terms are: T1 + T2 + T3 = 1 + 12x + 60x2 (ii) [(1 − x)(1 + 2x)]6 = (1 − x)6 (1 + 2x)6 Expanding up to x2 only [(1 − x)(1 + 2x)]6 = (1 − 6x + 15x2 )(1 + 12x + 60x2 ) Multiplying and neglecting x3 and higher powers [(1 − x)(1 + 2x)]6 = (1 + 12x + 60x2 − 6x − 72x2 + 15x2 Sum of coefficient of x2 =60-72+15=3 Hence coefficient of x2 in the expansion of [(1 − x)(1 + 2x)]6 =3 4