2014 A-level H2 Mathematics (9740) Paper 1 Question 6 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(a)
(i)
Let P(n) be the preposition, p_{n}=\frac{1}{3}(7-4^{n}) for n\in {\mathbb{Z}}^{+}
When n=1, \mathrm{LHS}=p_{1}=1 and \mathrm{RHS}=\frac{1}{3}(7-4)=1
Since \mathrm{LHS}=\mathrm{RHS}, P(1) is true.
Assume P(k) is true for some k \in {\mathbb{Z}}^{+}.
To prove that P(K+1) is true, i.e., p_{k+1}=\frac{1}{3}(7-4^{k+1})
\mathrm{LHS}
=p_{k+1}

=4p_{k}-7

=4(\frac{1}{3}(7-4^{k}))-7

= \frac{1}{3}(4\times{7}-4\times{4^{k}}-21)

= \frac{1}{3}(7-4^{k+1})
= \mathrm{RHS}
Therefore, P(k) is true \Rightarrow P(k+1) is also true.
Since P(1) is true and P(k) is true \Rightarrow P(k+1) is true, by principle of mathematical induction, P(n) is true for all n\in {\mathbb{Z}}^{+}.

(ii)
\sum_{r=1}^{n} p_{r}

= \sum_{r=1}^{n} {\frac{1}{3}}(7-4^{r})

= \frac{1}{3}[\sum_{r=1}^{n}7 - \sum_{r=1}^{n} 4^{r}]

= \frac{1}{3}\times{7n} - \frac{1}{3}[\frac{4(4^{n}-1)}{4-1}]

= \frac{7n}{3} - \frac{4}{9}(4^{n}-1)

(b)
(i)
As n\rightarrow \infty, \frac{1}{(n+1)!} \rightarrow 0
Therefore, the series converges and the \mathrm{Sum~to~Infinity}=1.

(ii)
u_{n}=S_{n}-S_{n-1}

u_{n}=1-\frac{1}{(n+1)!}-(1-\frac{1}{n!})

u_{n}=\frac{-1}{(n+1)!}+\frac{1}{n!}

u_{n}=\frac{-1+n+1}{(n+1)!}

u_{n}=\frac{n}{(n+1)!}

Personal Comments:
Firstly, students should know that (a) and (b) are unrelated, it is how A-level questions are denoted. For the MI, it is direct and a simple one on recurrence relation. We are then told to find the sum which tests us on our abilities to identify Arithmetic Progression and Geometric Progression in summation forms.
Convergence of a series can be proved in a few ways, the given is the most direct, which is to show that the sum to infinity is a finite value. Question wants students to write down, so please write it explicitly. It is surprising that some students confuse the formulas and write u_{n}=S_{n+1}-S_{n}, this is totally wrong.

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