2016 H2 Maths Paper 1 Q6 — Worked Solution

Part (i)

5 marks

Let $P(n)$ be the statement

\sum_{r = 1}^{n}r\left( r^{2} + 1 \right) = \frac{1}{4}n(n + 1)\left( n^{2} + n + 2 \right).

When $n = 1$ : LHS $\text{=}1\left( 1^{2} + 1 \right) = 2$ . RHS $\text{=}\frac{1}{4}(1)(2)(1 + 1 + 2) = \frac{1}{4}(2)(4) = 2$ . Since LHS $\text{=}$ RHS, $P(1)$ is true.

Assume $P(k)$ is true for some $k \in \mathbb{Z}^{\text{+}}$ , i.e.

\sum_{r = 1}^{k}r\left( r^{2} + 1 \right) = \frac{1}{4}k(k + 1)\left( k^{2} + k + 2 \right).

We want to show $P(k + 1)$ is true, i.e.

\sum_{r = 1}^{k + 1}r\left( r^{2} + 1 \right) = \frac{1}{4}(k + 1)(k + 2)\left( (k + 1)^{2} + (k + 1) + 2 \right).

\sum_{r = 1}^{k + 1}r\left( r^{2} + 1 \right)

= \sum_{r = 1}^{k}r\left( r^{2} + 1 \right) + (k + 1)\left( (k + 1)^{2} + 1 \right)

= \frac{1}{4}k(k + 1)\left( k^{2} + k + 2 \right) + (k + 1)\left( (k + 1)^{2} + 1 \right)

Factoring out $(k + 1)$ ,

\sum_{r = 1}^{k + 1}r\left( r^{2} + 1 \right) = (k + 1)\left\lbrack \frac{1}{4}k\left( k^{2} + k + 2 \right) + (k + 1)^{2} + 1 \right\rbrack

Simplifying the bracket,

\frac{1}{4}k\left( k^{2} + k + 2 \right) + (k + 1)^{2} + 1

= \frac{1}{4}\left( k^{3} + k^{2} + 2k \right) + k^{2} + 2k + 2

= \frac{k^{3} + k^{2} + 2k + 4k^{2} + 8k + 8}{4}

= \frac{k^{3} + 5k^{2} + 10k + 8}{4}

Since $(k + 2)\left( (k + 1)^{2} + (k + 1) + 2 \right) = (k + 2)\left( k^{2} + 3k + 4 \right) = k^{3} + 5k^{2} + 10k + 8$ , this bracket equals $\frac{1}{4}(k + 2)\left( (k + 1)^{2} + (k + 1) + 2 \right)$ , so

\sum_{r = 1}^{k + 1}r\left( r^{2} + 1 \right) = (k + 1) \cdot \frac{1}{4}(k + 2)\left( (k + 1)^{2} + (k + 1) + 2 \right)

= \frac{1}{4}(k + 1)(k + 2)\left( (k + 1)^{2} + (k + 1) + 2 \right).

This is $P(k + 1)$ , so $P(k)$ true $\Rightarrow$ $P(k + 1)$ true.

Since $P(1)$ is true, and $P(k)$ true $\Rightarrow P(k + 1)$ true, by the principle of mathematical induction $P(n)$ is true for all $n \in \mathbb{Z}^{\text{+}}$ . $\ \$ (shown)

Part (ii)

2 marks

Using the given recurrence $u_{n} = u_{n - 1} + n^{3} + n$ with $u_{0} = 2$ ,

u_{1} = u_{0} + 1^{3} + 1 = 2 + 2 = 4

u_{2} = u_{1} + 2^{3} + 2 = 4 + 10 = 14

u_{3} = u_{2} + 3^{3} + 3 = 14 + 30 = 44

$\therefore$ $u_{1} = 4$ , $u_{2} = 14$ , $u_{3} = 44$ .

Part (iii)

3 marks

Approach

The sum $\sum_{r = 1}^{n}\left( u_{r} - u_{r - 1} \right)$ telescopes to $u_{n} - u_{0}$ , since every interior term cancels. On the other hand, each term $u_{r} - u_{r - 1}$ equals $r^{3} + r$ by the given recurrence, and $r^{3} + r = r\left( r^{2} + 1 \right)$ , so this sum is exactly the sum evaluated in part (i). Equating the two gives $u_{n}$ .

Since $u_{r} - u_{r - 1} = r^{3} + r$ for $r \geq 1$ ,

\sum_{r = 1}^{n}\left( u_{r} - u_{r - 1} \right) = \sum_{r = 1}^{n}\left( r^{3} + r \right) = \sum_{r = 1}^{n}r\left( r^{2} + 1 \right)

By part (i), this equals $\frac{1}{4}n(n + 1)\left( n^{2} + n + 2 \right)$ .

Also, writing out the sum on the left, all terms cancel except the first and last:

\sum_{r = 1}^{n}\left( u_{r} - u_{r - 1} \right) = \left( u_{1} - u_{0} \right) + \left( u_{2} - u_{1} \right) + \ldots + \left( u_{n} - u_{n - 1} \right) = u_{n} - u_{0}

Comparing the two,

u_{n} - u_{0} = \frac{1}{4}n(n + 1)\left( n^{2} + n + 2 \right)

Since $u_{0} = 2$ ,

u_{n} = 2 + \frac{1}{4}n(n + 1)\left( n^{2} + n + 2 \right)

To check your answer: at

n = 3

,

u_{3} = 2 + \frac{1}{4}(3)(4)(9 + 3 + 2) = 2 + \frac{1}{4}(12)(14) = 2 + 42 = 44

, matching part (ii). ✓

$\therefore$ $u_{n} = 2 + \frac{1}{4}n(n + 1)\left( n^{2} + n + 2 \right)$ .

2016 H2 Maths Paper 1, Question 6

Part (i)

Part (ii)

Part (iii)

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