2023 H2 Maths Paper 1 Q2 — Worked Solution

Part (a)

3 marks

Approach

With four unknown coefficients and four data points, we set up four equations and eliminate the constant term $d$ , then the linear term $c$ , by taking successive differences, rather than solving the $4 \times 4$ system directly.

Let $u_{n} = an^{3} + bn^{2} + cn + d$ . Substituting $n = 1,2,3,4$ :

a + b + c + d = 10\quad\text{...(1)}

8a + 4b + 2c + d = 61\quad\text{...(2)}

27a + 9b + 3c + d = 206\quad\text{...(3)}

64a + 16b + 4c + d = 469\quad\text{...(4)}

Taking (2) $\text{-}$ (1), (3) $\text{-}$ (2) and (4) $\text{-}$ (3) eliminates $d$ :

7a + 3b + c = 51\quad\text{...(A)}

19a + 5b + c = 145\quad\text{...(B)}

37a + 7b + c = 263\quad\text{...(C)}

Taking differences again eliminates $c$ :

6a + b = 47\quad\text{...(D)}

9a + b = 59\quad\text{...(E)}

Subtracting (D) from (E):

3a = 12

\Rightarrow a = 4

Back-substituting: from (D), $b = 47 - 6(4) = 23$ ; from (A), $c = 51 - 7(4) - 3(23) = - 46$ ; from (1), $d = 10 - 4 - 23 - ( - 46) = 29$ .

To check your answer:

u_{1} = 4 + 23 - 46 + 29 = 10

✓,

u_{2} = 32 + 92 - 92 + 29 = 61

✓,

u_{3} = 108 + 207 - 138 + 29 = 206

✓,

u_{4} = 256 + 368 - 184 + 29 = 469

✓

$\therefore$ $u_{n} = 4n^{3} + 23n^{2} - 46n + 29$ .

Part (b)

2 marks

n \geq 17

Explanation: By GC, $u_{16} = 21\, 565$ and $u_{17} = 25\, 546$ , so $u_{16} < 25\, 000 < u_{17}$ . Since $u_{n}$ is a cubic with positive leading coefficient it is increasing for $n \geq 1$ , so once $u_{n}$ exceeds $25\, 000$ at $n = 17$ it stays above $25\, 000$ for every larger $n$ . As $n$ is a positive integer (a term index of the sequence), the range is $n \geq 17$ .

2023 H2 Maths Paper 1, Question 2

Part (a)

Part (b)

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