2017 H2 Maths Paper 1 Q9 — Worked Solution

Part (a)(i)

3 marks

Approach

$u_{n}$ is the $n^{\text{th}}$ term of the sequence, so $u_{n} = S_{n} - S_{n - 1}$ for $n \geq 2$ . We then check separately whether the resulting formula also holds at $n = 1$ .

For $n \geq 2$ ,

u_{n}

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

= \left( An^{2} + Bn \right) - \left( A(n - 1)^{2} + B(n - 1) \right)

= An^{2} + Bn - A\left( n^{2} - 2n + 1 \right) - Bn + B

= 2An - A + B

At $n = 1$ , $u_{1} = S_{1} = A + B$ , and the formula $2An - A + B$ at $n = 1$ gives $2A - A + B = A + B$ , which agrees. So the formula holds for all $n \geq 1$ .

$\therefore$ $u_{n} = 2An + B - A$ .

Part (a)(ii)

2 marks

Using the result of (i) with $n = 10$ and $n = 17$ :

19A + B = 48

33A + B = 90

Subtracting the first equation from the second:

14A = 42

A = 3

Substituting into $19A + B = 48$ :

19(3) + B = 48

B = - 9

To check your answer:

33(3) + ( - 9) = 99 - 9 = 90

, which matches. ✓

$\therefore$ $A = 3$ , $B = - 9$ .

Part (b)

4 marks

Approach

The left side is a difference of two squares in $(r + 1)$ and $(r - 1)$ , both multiplied by $r^{2}$ . Factorising this way avoids expanding everything out term by term.

r^{2}(r + 1)^{2} - (r - 1)^{2}r^{2}

= r^{2}\left\lbrack (r + 1)^{2} - (r - 1)^{2} \right\rbrack

= r^{2}\left\lbrack (r + 1 + r - 1)(r + 1 - r + 1) \right\rbrack

= r^{2}(2r)(2)

= 4r^{3}

$\therefore$ $k = 4$ . (shown)

Approach

With $f(r) = r^{2}(r + 1)^{2}$ , the result just shown says $f(r) - f(r - 1) = 4r^{3}$ . Summing this from $r = 1$ to $n$ telescopes, since every intermediate $f(r)$ cancels, leaving only the first and last terms.

Let $f(r) = r^{2}(r + 1)^{2}$ , so $f(r) - f(r - 1) = 4r^{3}$ .

\sum_{r = 1}^{n}4r^{3} = \sum_{r = 1}^{n}\left\lbrack f(r) - f(r - 1) \right\rbrack

Writing out the sum,

\sum_{r = 1}^{n}\left\lbrack f(r) - f(r - 1) \right\rbrack

= \left\lbrack f(1) - f(0) \right\rbrack + \left\lbrack f(2) - f(1) \right\rbrack + \ldots + \left\lbrack f(n) - f(n - 1) \right\rbrack

= f(n) - f(0)

= n^{2}(n + 1)^{2} - 0

So $4\sum_{r = 1}^{n}r^{3} = n^{2}(n + 1)^{2}$ , giving

\sum_{r = 1}^{n}r^{3} = \frac{n^{2}(n + 1)^{2}}{4}

To check your answer: at

n = 3

,

\sum r^{3} = 1 + 8 + 27 = 36

, and

\frac{3^{2}(4)^{2}}{4} = \frac{9 \cdot 16}{4} = 36

, which matches. ✓

$\therefore$ $\sum_{r = 1}^{n}r^{3} = \frac{n^{2}(n + 1)^{2}}{4}$ .

Part (c)

4 marks

Here $a_{r} = \frac{x^{r}}{r!}$ , so

\left| \frac{a_{n + 1}}{a_{n}} \right|

= \left| \frac{x^{n + 1}}{(n + 1)!} \cdot \frac{n!}{x^{n}} \right|

= \left| \frac{x}{n + 1} \right|

For any fixed real value of $x$ , the numerator $|x|$ is fixed while the denominator $n + 1 \rightarrow \infty$ as $n \rightarrow \infty$ , so

\lim_{n \rightarrow \infty}\left| \frac{a_{n + 1}}{a_{n}} \right| = 0

Since $0 < 1$ for every real $x$ (including $x = 0$ , where every term after $a_{0}$ is zero and the series trivially converges), the series converges for all real values of $x$ by the ratio test.

This series is the Maclaurin series of $e^{x}$ .

$\therefore$ $\sum_{r = 0}^{\infty}\frac{x^{r}}{r!}$ converges for all real $x$ , with sum to infinity $e^{x}$ .

2017 H2 Maths Paper 1, Question 9

Part (a)(i)

Part (a)(ii)

Part (b)

Part (c)

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