2024 H2 Maths Paper 2 Q4 — Worked Solution

Part (a)

2 marks

Since $S_{n} = t_{1} + t_{2} + \cdots + t_{n}$ , we have $t_{n} = S_{n} - S_{n - 1}$ for $n \geq 2$ , with $t_{1} = S_{1}$ checked separately.

S_{n - 1} = 2(n - 1)^{3} - 8(n - 1)^{2} - 4(n - 1)

= 2\left( n^{3} - 3n^{2} + 3n - 1 \right) - 8\left( n^{2} - 2n + 1 \right) - 4(n - 1)

= 2n^{3} - 14n^{2} + 18n - 6

Subtracting,

t_{n} = S_{n} - S_{n - 1}

= \left( 2n^{3} - 8n^{2} - 4n \right) - \left( 2n^{3} - 14n^{2} + 18n - 6 \right)

= 6n^{2} - 22n + 6

Note: this formula also holds at $n = 1$ , since $S_{1} = 2 - 8 - 4 = - 10$ agrees with $6 - 22 + 6 = - 10$ .

To check your answer:

t_{1} = 6(1)^{2} - 22(1) + 6 = - 10

, matching

S_{1} = 2(1)^{3} - 8(1)^{2} - 4(1) = - 10

✓

$\therefore$ $t_{n} = 6n^{2} - 22n + 6$ .

Part (b)

2 marks

50n - 204 = 6n^{2} - 22n + 6

\Rightarrow 6n^{2} - 72n + 210 = 0

\Rightarrow n^{2} - 12n + 35 = 0

\Rightarrow (n - 5)(n - 7) = 0

\Rightarrow n = 5\ \text{or}\ n = 7

$\therefore$ $n = 5$ or $n = 7$ .

Part (c)

2 marks

Approach

$U$ has common difference $50$ and $V$ has common difference $3$ , so a number common to both progressions recurs with period $lcm(50,3) = 150$ . Rather than listing terms of each series until one matches, we find one common term algebraically using modular arithmetic, restricting the search to terms exceeding $100$ .

Terms of $U$ exceed $100$ when

50n - 204 > 100

\Rightarrow n > 6.08

\Rightarrow n \geq 7,

giving $u_{7} = 146,\ u_{8} = 196,\ u_{9} = 246,\ldots$

For a term to lie in both series, we need $50a - 204 = 3b + 16$ for positive integers $a,b$ :

50a - 204 = 3b + 16

\Rightarrow 50a - 220 = 3b

Taking both sides mod 3 (using $50 \equiv 2$ and $220 \equiv 1$ ),

2a \equiv 1\mkern6mu \left( \text{mod }3 \right)

\Rightarrow a \equiv 2\mkern6mu \left( \text{mod }3 \right),

since $2$ is its own inverse mod 3. So $a = 2,5,8,11,\ldots$ . The smallest such $a$ with $a \geq 7$ (so that $u_{a} > 100$ ) is $a = 8$ :

u_{8} = 50(8) - 204 = 196

To check your answer:

196 = 3b + 16 \Rightarrow b = 60

, an integer, so

196 = v_{60}

✓

$\therefore$ the smallest number greater than 100 in both series is $196$ .

Part (d)(i)

2 marks

Approach

since $w_{n}$ depends on the parity of $n$ , we check $n$ even and $n$ odd separately, writing $n = 2k$ or $n = 2k + 1$ .

If $n = 2k$ ,

w_{n} = 3(2k)^{2} - 5(2k) + 7

= 12k^{2} - 10k + 7

$12k^{2}$ and $10k$ are both even, so $w_{n} = even - even + odd = odd$ .

If $n = 2k + 1$ ,

w_{n} = 3(2k + 1)^{2} - 5(2k + 1) + 7

= 3\left( 4k^{2} + 4k + 1 \right) - 5(2k + 1) + 7

= 12k^{2} + 2k + 5

$12k^{2}$ and $2k$ are both even, so $w_{n} = even + even + odd = odd$ .

$\therefore$ in both cases $w_{n}$ is odd, so every term of series $W$ is odd. (shown)

Part (d)(ii)

1 mark

u_{n} = 50n - 204

$50n$ is even for every integer $n$ , and $204$ is even, so $u_{n} = even - even = even$ for every $n$ — every term of $U$ is even.

Since every term of $W$ is odd (part (i)) and every term of $U$ is even, no term of $W$ can equal a term of $U$ .

$\therefore$ series $U$ and series $W$ have no terms in common.

2024 H2 Maths Paper 2, Question 4

Part (a)

Part (b)

Part (c)

Part (d)(i)

Part (d)(ii)

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