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N2023 P1 Q2

N2023 P1 Q2

Junior College 2
5 marks
Free

The first four terms of a sequence are given by \(u_1=10\), \(u_2=61\), \(u_3=206\) and \(u_4=469\). It is given that \(u_n\) is a cubic polynomial.

  1. Find \(u_n\) in terms of \(n\).[3]
  2. Find the range of values of \(n\) for which \(u_n\) is greater than \(25\,000\).[2]

Default solution

  1. Let \(u_n=an^3+bn^2+cn+d\). Substituting the four given terms:
    \(a+b+c+d\)\({}=10\)\(8a+4b+2c+d\)\({}=61\)\(27a+9b+3c+d\)\({}=206\)\(64a+16b+4c+d\)\({}=469\)
    Subtract consecutive equations to eliminate \(d\):
    \(7a+3b+c\)\({}=51\)\(19a+5b+c\)\({}=145\)\(37a+7b+c\)\({}=263\)
    Subtract again to eliminate \(c\):
    \(12a+2b\)\({}=94\)\(18a+2b\)\({}=118\)\(6a\)\({}=24\)Subtract the two equations.\(a\)\({}=4\)\(b\)\({}=23\)\(c\)\({}=51-7(4)-3(23)=-46\)\(d\)\({}=10-4-23+46=29\)
    \(\therefore u_n=4n^3+23n^2-46n+29.\)
  2. \(u_{n+1}-u_n\)\({}=12n^2+58n-19\)\(u_{n+1}-u_n\)\({}\ge12+58-19=51>0\)
    The sequence is strictly increasing for positive integer \(n\). A calculator table at the threshold gives
    \(n\)\(u_n\)
    \(16\)\(21565\)
    \(17\)\(25546\)
    \(\therefore n\ge17,\qquad n\in\mathbb Z.\)
Answer:(a) \(u_n=4n^3+23n^2-46n+29\) (b) \(n\ge17\), \(n\in\mathbb Z\)

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