H2 Math Sun 2pm

This page contains all questions and answers asked by students from this class. The most recent questions will be at the top.

MF26

Question 15 CJC/2013

$LHS = \sum_{r=2}^n \frac{1}{(r+1)(r-1)}$
$= \sum_{r=2}^n \frac{1}{2} (\frac{1}{r-1} - \frac{1}{r+1})$
$= \frac{1}{2} \sum_{r=2}^n \frac{1}{r-1} - \frac{1}{r+1}$
$= \frac{1}{2} [\frac{1}{1} - \frac{1}{3}$
$+ \frac{1}{2} - \frac{1}{4}$
$+ \frac{1}{3} - \frac{1}{5}$
…
$+ \frac{1}{n-3} - \frac{1}{n-1}$
$+ \frac{1}{n-2} - \frac{1}{n}$
$+ \frac{1}{n-1} - \frac{1}{n+1}]$
$= \frac{1}{2} [\frac{3}{2} - \frac{1}{n} - \frac{1}{n+1}]$
$= \frac{3}{4} - \frac{1}{2n} - \frac{1}{2n+2}$

Since the $lim_{n \rightarrow \infty} \sum_{r=2}^n \frac{1}{(r+1)(r-1)}$
$= lim_{n \rightarrow \infty} \frac{3}{4} - \frac{1}{2n} - \frac{1}{2n+2}$
$= \frac{3}{4} - 0 - 0$
$- \frac{3}{4}$ is a constant, the series convergences.
The sum to infinity $= \frac{3}{4}$

Question 14a MJC/2013

Let $T_n$ denote the $n^{th}$ term of the AP.
$T_n = a + (n-1)d$
$T_1 = a$
$T_3 = a +2d$
$T_6 = a +5d$
Since they are consecutive terms of a GP,
$\frac{T_3}{T_1} = \frac{T_6}{T_3} = r$
$\Rightarrow \frac{a+2d}{a} = \frac{a+5d}{a+2d}$
$(a+2d)^2 = a(a+5d)$
$a^2 + 4ad + 4d^2 = a^2 + 5ad$
$4d^2 - ad =0$
$d(4d - a) = 0$
$d = 0 (NA) \mathrm{~or~} 4d = a$
$\Rightarrow r = \frac{a+2d}{a}$
$r = \frac{6d+2d}{4d} = \frac{3}{2} > 1$ , thus its not convergent

$S_{15} = \frac{n}{2}(2a + (n-1)d)$
$= \frac{15}{2}(2a + 14 (\frac{a}{4}))$
$= 41.25 a$

Question 11 DHS/2013

Sum of first 3 terms $= \frac{3}{2} (2a+(3-1)d)$
$6 = 3a+3d$
$a+d=2$ —(1)
Sum of last 3 terms $= \frac{3}{2}[2(a+(n-1)d) + (3-1)(-d)]$ ; Here we consider an AP that has first term $T_n = a + (n-1)d$ and common difference $-d$ .
$\Rightarrow 231 = \frac{3}{2}(2a + 2nd - 2d -2d)$
$231 = 3a + 3nd - 6d$
$77 = a + nd -2d$ —(2)
Sum of n terms = $\frac{n}{2}[2a+ (n-1)d]$
$1106 = \frac{n}{2}(2a + nd - d)$ —(3)
Solve for n.

Question 12 SRJC/2012

(i) Volume, $V = \pi r^2 h$
Volume of kth later, $V_k = \pi [(20)(\frac{5}{6})^{k-1}]^2(22)(\frac{4}{5})^{k-1}$
$V_k = 8800 \pi [\frac{25}{36} \times \frac{4}{5}]^{k-1}$
$V_k = 8800 \pi (\frac{5}{9})^{k-1}$
(ii)
Since $r = \frac{5}{9} <1, S_{\infty}$ exists.
Theoretical Max Volume, $S_{\infty} = \frac{8800 \pi}{1 - \frac{5}{9}} = 19800 \pi$ .
Total Volume, $S_n = \frac{8800 \pi (1 - (\frac{5}{9})^n)}{1 - \frac{5}{9}}$
We want $S_n \le 0.95 S_{\infty}$

About

KS Teng

KS has been teaching H2/H1 Mathematics and IB mathematics for the more than 10 years. Having taught students from all various Junior Colleges, KS adapts to students' abilities and help them better understand the topics. As someone who loves to teach mathematics and sees it as a truly useful tool in life, KS seeks to enable students to appreciate math. Therefore, his tuition mission is to motivate and cultivate students to be independent and confident thinkers.

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