By R. Haines, B. Haines
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Extra resources for Work Out Pure Mathematics A-Level
Find the values of a and d. 21 • Fifth term of an arithmetic progression: a + 4d = 59. Sum of first 30 terms = 30(2a + 29d)/2 = 30a + 435d. Sum of first 10 terms = 1O(2a + 9d)/2 = lOa + 45d. Hence 30a + 435d = 4(10a + 45d). :. lOa = 255d ~ 2a = SId, From equations 1 and 2, a = 51 and d = 2. 3 The first three terms of a geometric progression are 6, t,-f-;. 75. 7499 . 1. e. t 6 Sum to infinity S oc ~ r = t. 75. 75(t)n. 0611. Hence the least number of terms required is 6. 4 (a) An arithmetic progression is such that the sum of the first n terms is 2n 2 for all positive integral values of n.
1 = 1 + x + x2 + x 3 (1 - X)-1 Hence -\1(1 + x) _ for Ix I < 1. +... 2 ll. 3 - 1 + 2X + 8 X + 16X + .... (1 - x) When x --.!. 1. for Ixl
L +x) (l - 2x) Hence, or otherwise, find the first three terms in the expansion of the function in ascending powers of x. 6 Express the function • Notice that one of the factors is a repeated factor => there must be three fractions, denominators (1 + x), (1 - 2x) and (1 - 2x)2 . - 2x) (1 - 2X)2 (1 +x) (1 - 2x)2 _ A(1 - 2X)2 +B(1 +x) (1 - 2x) + C(l +x) = (1 + x) (1 - 2X)2 . Equating the numerators: 9x =A(1 - 2x)2 + B(1 + x) (1 - 2x) + C(1 + x). -9 =A(3)2 => A =-l. t=C(t) => C=3. O=A+B+C => B=-2.
Work Out Pure Mathematics A-Level by R. Haines, B. Haines