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

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

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

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

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}

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