2018 H2 Maths Paper 1 Q8 — Worked Solution

Part (i)

2 marks

Since $u_{n + 1} = 2u_{n} + An$ , taking $n = 1$ with $u_{1} = 5$ and $u_{2} = 15$ ,

u_{2} = 2u_{1} + A(1)

15 = 10 + A

A = 5.

Taking $n = 2$ ,

u_{3} = 2u_{2} + A(2) = 2(15) + 5(2) = 40.

$\therefore A = 5$ and $u_{3} = 40$ .

Part (ii)

4 marks

Approach

We are told $u_{n} = a\left( 2^{n} \right) + bn + c$ satisfies the same recurrence as $u_{n + 1} = 2u_{n} + 5n$ (using $A = 5$ from (i)). Substituting this form into the recurrence and comparing coefficients of $n$ on both sides — since the equation must hold for every $n$ — gives $b$ and $c$ ; $a$ is then found from the initial condition $u_{1} = 5$ , since the recurrence alone cannot fix $a$ (any multiple of the solution of the homogeneous equation $u_{n + 1} = 2u_{n}$ can be added).

Substituting $u_{n} = a\left( 2^{n} \right) + bn + c$ into $u_{n + 1} = 2u_{n} + 5n$ ,

a\left( 2^{n + 1} \right) + b(n + 1) + c = 2\left\lbrack a\left( 2^{n} \right) + bn + c \right\rbrack + 5n

a\left( 2^{n + 1} \right) + bn + b + c = a\left( 2^{n + 1} \right) + 2bn + 2c + 5n

bn + b + c = (2b + 5)n + 2c

0 = (b + 5)n + (c - b).

Since this holds for every $n$ , both coefficients vanish:

b + 5 = 0 \Longrightarrow b = - 5,

c - b = 0 \Longrightarrow c = b = - 5.

Using $u_{1} = 5$ with $b = - 5$ , $c = - 5$ ,

u_{1} = a(2) + b(1) + c

5 = 2a - 5 - 5

2a = 15

a = \frac{15}{2}.

To check your answer: with

a = \frac{15}{2}

,

b = - 5

,

c = - 5

:

u_{2} = \frac{15}{2}(4) - 10 - 5 = 15

and

u_{3} = \frac{15}{2}(8) - 15 - 5 = 40

, matching (i). ✓

$\therefore a = \frac{15}{2}$ , $b = - 5$ , $c = - 5$ .

Part (iii)

4 marks

Since $u_{r} = \frac{15}{2}\left( 2^{r} \right) - 5r - 5$ ,

\sum_{r = 1}^{n}u_{r} = \frac{15}{2}\sum_{r = 1}^{n}2^{r} - 5\sum_{r = 1}^{n}r - 5\sum_{r = 1}^{n}1.

By GC (or the standard GP and AP sum formulas), $\sum_{r = 1}^{n}2^{r} = \frac{2\left( 2^{n} - 1 \right)}{2 - 1} = 2^{n + 1} - 2$ and $\sum_{r = 1}^{n}r = \frac{n(n + 1)}{2}$ , so

\sum_{r = 1}^{n}u_{r} = \frac{15}{2}\left( 2^{n + 1} - 2 \right) - 5 \cdot \frac{n(n + 1)}{2} - 5n.

$\therefore\sum_{r = 1}^{n}u_{r} = \frac{15}{2}\left( 2^{n + 1} - 2 \right) - \frac{5n(n + 1)}{2} - 5n$ .

2018 H2 Maths Paper 1, Question 8

Part (i)

Part (ii)

Part (iii)

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