1998_Paper II

# 1998_Paper II - Section A 1 Pure Mathematics Show that if n...

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Section A: Pure Mathematics 1 Show that, if n is an integer such that ( n - 3) 3 + n 3 = ( n + 3) 3 , ( * ) then n is even and n 2 is a factor of 54. Deduce that there is no integer n which satisfies the equation ( * ). Show that, if n is an integer such that ( n - 6) 3 + n 3 = ( n + 6) 3 , ( ** ) then n is even. Deduce that there is no integer n which satisfies the equation ( ** ). 2 Use the first four terms of the binomial expansion of (1 - 1 / 50) 1 / 2 , writing 1 / 50 = 2 / 100 to simplify the calculation, to derive the approximation 2 1 . 414214. Calculate similarly an approximation to the cube root of 2 to six decimal places by considering (1 + N/ 125) a , where a and N are suitable numbers. [You need not justify the accuracy of your approximations.] 3 Show that the sum S N of the first N terms of the series 1 1 . 2 . 3 + 3 2 . 3 . 4 + 5 3 . 4 . 5 + · · · + 2 n - 1 n ( n + 1)( n + 2) + · · · is 1 2 3 2 + 1 N + 1 - 5 N + 2 . What is the limit of S N as N → ∞ ? The numbers a n are such that a n a n - 1 = ( n - 1)(2 n - 1) ( n + 2)(2 n - 3) . Find an expression for a n /a 1 and hence, or otherwise, evaluate n =1 a n when a 1 = 2 9 . 1

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4 The integral I n is defined by I n = Z π 0 ( π/ 2 - x ) sin( nx + x/ 2) cosec ( x/ 2) d x, where n is a positive integer. Evaluate I n - I n - 1 , and hence evaluate I n leaving your answer in the form of a sum. 5 Define the modulus of a complex number z and give the geometric interpretation of | z 1 - z 2 | for two complex numbers z 1 and z 2 . On the basis of this interpretation establish the inequality | z 1 + z 2 | 6 | z 1 | + | z 2 | . Use this result to prove, by induction, the corresponding inequality for | z 1 + · · · + z n | . The complex numbers a 1 , a 2 , . . . , a n satisfy | a i | 6 3 ( i = 1 , 2 , . . . , n ). Prove that the equation a 1 z + a 2 z 2 · · · + a n z n = 1 has no solution z with | z | 6 1 / 4.
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