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a) Use mathematical induction to prove that for all integers n ≥ 3, \( \left(1 - \frac{2}{3}\right)\left(1 - \frac{2}{4}\right)\left(1 - \frac{2}{5}\right) \cdots \left(1 - \frac{2}{n}\right) = \frac{2}{n(n-1)} - HSC - SSCE Mathematics Extension 1 - Question 4 - 2004 - Paper 1
Question 4
a) Use mathematical induction to prove that for all integers n ≥ 3, \( \left(1 - \frac{2}{3}\right)\left(1 - \frac{2}{4}\right)\left(1 - \frac{2}{5}\right) \cdots \... show full transcript
Worked Solution & Example Answer:a) Use mathematical induction to prove that for all integers n ≥ 3, \( \left(1 - \frac{2}{3}\right)\left(1 - \frac{2}{4}\right)\left(1 - \frac{2}{5}\right) \cdots \left(1 - \frac{2}{n}\right) = \frac{2}{n(n-1)} - HSC - SSCE Mathematics Extension 1 - Question 4 - 2004 - Paper 1
Step 1
a) Use mathematical induction to prove that for all integers n ≥ 3, ...
Answer
To prove this statement using mathematical induction, we follow these steps:
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Base Case (n = 3): Start with n = 3: [ \left(1 - \frac{2}{3}\right) = \frac{1}{3} ] The right-hand side is: [ \frac{2}{3(3-1)} = \frac{2}{6} = \frac{1}{3} ] The base case holds.
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Inductive Step: Assume true for n = k: [ \left(1 - \frac{2}{3}\right)\left(1 - \frac{2}{4}\right)...\left(1 - \frac{2}{k}\right) = \frac{2}{k(k-1)} ] Need to show for n = k + 1: [ \left(1 - \frac{2}{3}\right)\left(1 - \frac{2}{4}\right)...\left(1 - \frac{2}{k}\right) \left(1 - \frac{2}{k + 1}\right) ] Which simplifies to: [ \frac{2}{k(k-1)} \left(1 - \frac{2}{k+1}\right) = \frac{2}{k(k-1)} \cdot \frac{k - 1}{k + 1} ] Therefore: [ = \frac{2}{(k+1)k} ] Thus, by induction, the statement holds for all n ≥ 3.
Step 2
b) (i) The equation of the tangent to x^2 = 4ay ...
Answer
The tangents to the points P and Q can be expressed using:
- For point P: ( y = \frac{p}{2a} x - ap^2 )
- For point Q: ( y = \frac{q}{2a} x - aq^2 )
Setting these two equations equal gives: [ \frac{p}{2a} x - ap^2 = \frac{q}{2a} x - aq^2 ] Rearranging leads to: [ \left(\frac{p - q}{2a}\right) x = a(p^2 - q^2) ] Solving for x enables us to find R, namely ( R = (a(p + q), apq) ).
Step 3
b) (ii) Find the locus of R.
Answer
Since ( \angle POQ ) is always a right angle, we can use the fact that the slopes of the lines from O to P and Q must yield: [ \frac{ap - 0}{2a(\frac{p}{2a})} \cdot \frac{aq-0}{2a(\frac{q}{2a})} = -1 ] Simplifying gives us the condition necessary to determine the locus of R, leading to the expression: [ y = ax(1 - \frac{pq}{2}) \text{ (in parametric form)} ]
Step 4
c) (i) What is the probability ...
Answer
The probability Katie wins at least 1 prize in 7 draws can be calculated using the complement: [ P(\text{at least 1}) = 1 - P(\text{none}) ] Here, the probability she does not win in one draw is (\frac{9}{10}), so for 7 draws: [ P(\text{none}) = \left(\frac{9}{10}\right)^7 ] Hence: [ P(\text{at least 1}) = 1 - \left(\frac{9}{10}\right)^7 ]
Step 5
c) (ii) Show that in the first 20 weeks ...
Answer
To compare the probabilities of winning exactly 1 and 2 prizes: The probability of winning exactly k prizes in n draws follows a binomial distribution: [ P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k} ] Where p = \frac{1}{10}. Specifically:
- For 1 prize: [ P(X = 1) = \binom{20}{1} \left(\frac{1}{10}\right)^1 \left(\frac{9}{10}\right)^{19} ]
- For 2 prizes: [ P(X = 2) = \binom{20}{2} \left(\frac{1}{10}\right)^2 \left(\frac{9}{10}\right)^{18} ] By comparing these two expressions, we can show the greater chance of exactly 2 wins.
Step 6
c) (iii) Most candidates ...
Answer
For part (iii), the optimal parameters for Katie's participation can be calculated through the probability of winning exactly 3 prizes and exactly 2 prizes:
- Using binomial probability formulas: [ P(X = 3) = \binom{n}{3} p^3 (1-p)^{n-3}\text{ and } P(X = 2) = \binom{n}{2} p^2 (1-p)^{n-2} ] To find the specific parameters that would provide a better chance for winning exactly 3 prizes compared to 2 depends on evaluating these expressions with various n and p.