2020 H2 Maths Paper 2 Q2 — Worked Solution

Part (a)

Approach

Parts (i) and (ii) all concern the same recurrence $u_{n + 1} = 2u_{n} - 5$ . Rather than generate terms one at a time, we look for the fixed point of the recurrence — the value $u$ for which $u = 2u - 5$ — and write $u_{n} - u$ as a geometric sequence about it. This gives a closed form we can reuse for every part.

Since the fixed point satisfies $u = 2u - 5$ , we get $u = 5$ . Subtracting this from the recurrence,

u_{n + 1} - 5 = 2u_{n} - 10

= 2\left( u_{n} - 5 \right).

So $u_{n} - 5$ is geometric with common ratio $2$ and first term $u_{1} - 5 = p - 5$ , giving

u_{n} = 5 + (p - 5) \cdot 2^{n - 1}.

Part (i)

Part (A)

1 mark

With $p = 7$ ,

u_{n} = 5 + 2 \cdot 2^{n - 1}

= 5 + 2^{n}.

Explanation: since $2^{n} \rightarrow \infty$ as $n \rightarrow \infty$ , the terms increase strictly and without bound; the sequence diverges to infinity.

Part (B)

1 mark

With $p = 5$ ,

u_{n} = 5 + 0 \cdot 2^{n - 1} = 5.

Explanation: every term equals $5$ ; the sequence is constant.

Part (ii)

2 marks

Using the formula with $n = 5$ :

u_{5} = 5 + (p - 5) \cdot 2^{4}

= 5 + 16(p - 5).

Using the given value $u_{5} = 101$ :

5 + 16(p - 5) = 101

16(p - 5) = 96

p - 5 = 6

p = 11.

$\therefore p = 11$ .

Part (b)

Approach

We need $v_{3}$ and $v_{4}$ in terms of $a$ and $b$ before we can use the condition $v_{4} = 2v_{3}$ .

Using the recurrence with $n = 1$ :

v_{3} = v_{1} + 2v_{2} - 7 = a + 2b - 7.

Using the recurrence with $n = 2$ :

v_{4} = v_{2} + 2v_{3} - 7

= b + 2(a + 2b - 7) - 7

= 2a + 5b - 21.

Part (i)

3 marks

Substituting into $v_{4} = 2v_{3}$ :

2a + 5b - 21 = 2(a + 2b - 7)

2a + 5b - 21 = 2a + 4b - 14

b = 7.

$\therefore b = 7$ .

Part (ii)

1 mark

With $b = 7$ ,

v_{3} = a + 2(7) - 7 = a + 7,

v_{4} = 2v_{3} = 2a + 14.

Using the recurrence with $n = 3$ :

v_{5} = v_{3} + 2v_{4} - 7

= (a + 7) + 2(2a + 14) - 7

= 5a + 28.

$\therefore v_{5} = 5a + 28$ .

Part (c)

Part (i)

2 marks

Since $S_{n} = n^{3} - 11n^{2} + 4n$ is the sum of the first $n$ terms, the $n$ th term is $T_{n} = S_{n} - S_{n - 1}$ for $n > 1$ .

T_{n} = \left( n^{3} - 11n^{2} + 4n \right) - \left\lbrack (n - 1)^{3} - 11(n - 1)^{2} + 4(n - 1) \right\rbrack

= 3n^{2} - 25n + 16.

Note: checking $n = 1$ , $T_{1} = 3 - 25 + 16 = - 6 = S_{1}$ , so this formula also gives the first term correctly and holds for all $n \geq 1$ .

$\therefore T_{n} = 3n^{2} - 25n + 16$ .

Part (ii)

2 marks

Approach

“The sum of the first $m$ terms equals the sum of the first three terms” means $S_{m} = S_{3}$ , so $n = m$ and $n = 3$ are both roots of $S_{n} = S_{3}$ ; we factor out $(n - 3)$ to find the others.

S_{3} = 3^{3} - 11(3)^{2} + 4(3) = - 60.

Setting $S_{m} = - 60$ :

m^{3} - 11m^{2} + 4m = - 60

m^{3} - 11m^{2} + 4m + 60 = 0.

Since $m = 3$ is a root, we factor it out:

(m - 3)\left( m^{2} - 8m - 20 \right) = 0

(m - 3)(m - 10)(m + 2) = 0

m = 3,\ 10,\ - 2.

Since $m > 3$ , $m = 3$ and $m = - 2$ are rejected $\left( m = - 2\ \text{also rejected as}\ m\ \text{is a positive integer} \right)$ .

To check your answer:

S_{10} = 1000 - 1100 + 40 = - 60 = S_{3}

. ✓

$\therefore m = 10$ .

2020 H2 Maths Paper 2, Question 2

Part (a)

Part (i)

Part (A)

Part (B)

Part (ii)

Part (b)

Part (i)

Part (ii)

Part (c)

Part (i)

Part (ii)

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