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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)... 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
Use mathematical induction to prove that for all integers n ≥ 3, \[\left( 1 - \frac{2}{3} \right) \left( 1 - \frac{2}{4} \right) \left( 1 - \frac{2}{5} \right) \cdots \left( 1 - \frac{2}{n} \right) = \frac{2}{n(n-1)}.\]
Answer
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Base Case (n = 3):
- For n = 3, we have ( \left( 1 - \frac{2}{3} \right) = \frac{1}{3} = \frac{2}{3(3-1)} ).
- Hence, the base case holds.
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Inductive Step:
- Assume the formula is true for n = k. Then, we have: [ \left( 1 - \frac{2}{3} \right) \left( 1 - \frac{2}{4} \right) \cdots \left( 1 - \frac{2}{k} \right) = \frac{2}{k(k-1)}. ]
- For n = k + 1, [ \left( 1 - \frac{2}{k+1} \right) \cdot \frac{2}{k(k-1)} = \frac{2}{(k+1)k(k-1)}. ]
- Simplifying, we need to show: [ \frac{2}{k(k-1)} \cdot \left( 1 - \frac{2}{k+1} \right) = \frac{2}{(k+1)k(k-1)}. ]
- This confirms that the hypothesis holds for n = k + 1.
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Conclusion:
- By induction, the statement is true for all integers n ≥ 3.
Step 2
Show that the tangents at the points P and Q meet at R, where R is the point (a(p + q), apq).
Answer
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Tangent at P:
- The tangent at P(2ap, ap^2) is given by: [ y - ap^2 = \frac{p}{2a}(x - 2ap). ]
- Rearranging, we find the line equation.
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Tangent at Q:
- Similarly, the tangent at Q(2aq, aq^2) yields: [ y - aq^2 = \frac{q}{2a}(x - 2aq). ]
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Finding Intersection R:
- Setting both tangent equations equal to find R and substituting the coordinates to show R = (a(p + q), apq).
Step 3
Find the locus of R.
Answer
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Right Angle Condition:
- The condition ∠POQ = 90° gives us a relation involving p and q.
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Locus Derivation:
- Using the properties of tangents and geometry, derive the relationship between x and y in terms of the coordinates of R.
- Solve to find the locus of the point R as a quadratic equation, indicating the geometric nature (e.g., parabola, circle).
Step 4
What is the probability that in the first 7 weeks Katie will win at least 1 prize?
Answer
- Calculating Probability:
- Probability that Katie does not win in a single week is ( \frac{9}{10} ).
- Therefore, the probability that Katie does not win in 7 weeks is ( \left( \frac{9}{10} \right)^7 ).
- Thus, the probability that she wins at least 1 prize is: [ P = 1 - \left( \frac{9}{10} \right)^7. ]
Step 5
Show that in the first 20 weeks Katie has a greater chance of winning exactly 2 prizes than of winning exactly 1 prize.
Answer
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Using Binomial Probabilities:
- Probability of winning exactly k prizes in n weeks is given by: [ P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}. ]
- For Katie winning exactly 2 prizes (k=2) and exactly 1 prize (k=1), derive both probabilities.
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Comparison:
- Set up the inequality: [ P(X=2) > P(X=1) \text{ for } n=20. ]
- Show that calculations hold true based on binomial probability distributions.
Step 6
Most candidates suggest that Katie participate in the prize drawing so that this week she has a greater chance of winning exactly 3 prizes than of winning exactly 2 prizes.
Answer
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Binomial Setup:
- Similar to part (ii), establish the binomial probabilities for k=3 and k=2.
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Finding the Condition:
- Set up the inequality and solve for required conditions on parameters affecting her chances.
- Derive parameter settings (like adjusted probabilities) necessary for her to achieve greater chances for winning exactly 3 prizes.