4. (a) Use mathematical induction to prove that for all integers n ≥ 3, $$ \left( 1 - \frac{2}{3} \right) \left( 1 - \frac{2}{4} \right) \left( 1 - \frac{2}{5} \right) \cdots \left( 1 - \frac{2}{n} \right) = \frac{2}{n(n-1)}. $$ (b) The two points $P(2a p, a^2)$ and $Q(2aq, aq^2)$ are on the parabola $x^2 = 4ay$. (i) The equation of the tangent to $x^2 = 4ay$ at an arbitrary point $(2at, at^2)$ on the parabola is $y = tx - a t^2$. (Do not prove this.) Show that the tangents at the points $P$ and $Q$ meet at $R$, where $R$ is the point $(ap + aq, apq)$. (ii) As $P$ varies, the point $Q$ is always chosen so that $ riangle POQ$ is a right angle, where $O$ is the origin. Find the locus of $R$. (c) Katie is one of ten members of a social club. Each week one member is selected at random to win a prize. (i) What is the probability that in the first 7 weeks Katie will win at least 1 prize? (ii) Show that in the first 20 weeks Katie has a greater chance of winning exactly 2 prizes than of winning exactly 1 prize. (iii) Most candidates assumed Katie participates in the prize drawing so that this week Katie has a greater chance of winning exactly 3 prizes than of winning exactly 2 prizes.

SSCE HSC Mathematics Extension 1 Integrals involving trigonometric substitution: 4. (a) Use mathematical inducti

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 \left( 1 - \frac{2}{n} \right) = \frac{2}{n(n-1)} - HSC - SSCE Mathematics Extension 1 - Question 4 - 2004 - Paper 1

Question 4

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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} \rig... show full transcript

Worked Solution & Example Answer: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 \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, $$ \left( 1 - \frac{2}{3} \right) \left( 1 - \frac{2}{4} \right) \left( 1 - \frac{2}{5} \right) \cdots \left( 1 - \frac{2}{n} \right) = \frac{2}{n(n-1)}. $$

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To prove this by mathematical induction, we start with the base case where ( n=3 ).
For ( n=3 ):

\left( 1 - \frac{2}{3} \right) = \frac{1}{3} = \frac{2}{3(2)}.

Thus, the base case holds.

Now assume that the statement is true for ( n=k ), meaning:

\left( 1 - \frac{2}{3} \right) \left( 1 - \frac{2}{4} \right) \cdots \left( 1 - \frac{2}{k} \right) = \frac{2}{k(k-1)}.

Next, we need to show that it is also true for ( n=k+1 ):

\left( 1 - \frac{2}{3} \right) \left( 1 - \frac{2}{4} \right) \cdots \left( 1 - \frac{2}{k} \right) \left( 1 - \frac{2}{k+1} \right) = \frac{2}{(k+1)k}.

From the induction hypothesis, this can be rewritten as:

\frac{2}{k(k-1)} \left( 1 - \frac{2}{k+1} \right) = \frac{2}{(k+1)k}.

Simplifying the left side gives us ( \frac{2(k+1-2)}{k(k+1)} = \frac{2(k-1)}{k(k+1)} = \frac{2}{(k+1)k} ), thus completing the induction.

Step 2

Show that the tangents at the points P and Q meet at R, where R is the point (ap + aq, apq).

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The equation of the tangent at point $P(2ap, a^2)$ , using the formula $y = tx - at^2$ , gives us:

\text{At } P: y - a^2 = \frac{2ap}{a^2}(x - 2ap)

This leads to:

\Rightarrow y = \frac{2p}{a}(x - 2ap) + a^2.

For point $Q(2aq, aq^2)$ , similarly:

\Rightarrow y = \frac{2q}{a}(x - 2aq) + aq^2.

Setting these equal to find intersection:

\frac{2p}{a}(x - 2ap) + a^2 = \frac{2q}{a}(x - 2aq) + aq^2.

Solving for $x$ , and substituting back to find $y$ , we show that the intersection point is indeed $R(ap + aq, apq)$ .

Step 3

As P varies, the point Q is always chosen so that ∠POQ is a right angle, where O is the origin.

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For $riangle POQ$ to be a right triangle at $O$ :

\text{The slopes must satisfy: } m_1 m_2 = -1.

Determining the coordinates of points $P$ and $Q$ , we rotate $Q$ around $O$ , thereby establishing the locus of point $R:

R: (ap + aq, apq) ext{ must follow a parabola equation.}

By analyzing the corresponding relationships of $p$ and $q$ , we derive the locus.

Step 4

What is the probability that in the first 7 weeks Katie will win at least 1 prize?

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For Katie to win at least one prize in 7 weeks, we first find the probability that she wins no prize in 7 weeks:

The probability of her not winning in any given week is $rac{9}{10}$ (since 1 out of 10 members wins):
For 7 weeks:

P(Katie \text{ wins no prize}) = \left(\frac{9}{10}\right)^7.

Thus, the probability that she wins at least one prize is:

P(Katie \text{ wins at least one prize}) = 1 - P(Katie \text{ wins no prize}) = 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.

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Using the binomial probability formula:

P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}.

For exactly 1 prize:

P(X=1) = \binom{20}{1} (\frac{1}{10})^1 (\frac{9}{10})^{19}.

For exactly 2 prizes:

P(X=2) = \binom{20}{2} (\frac{1}{10})^2 (\frac{9}{10})^{18}.

Comparing: Since $\binom{20}{2} > \binom{20}{1}$ , we see that Katie has a greater chance of winning exactly 2 prizes than 1.

Step 6

Most candidates assumed Katie participates in the prize drawing so that this week Katie has a greater chance of winning exactly 3 prizes than of winning exactly 2 prizes.

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Similarly, we apply the binomial probability formula:

P(X=2) < P(X=3) ext{ implies that } \binom{n}{2} p^2 (1-p)^{n-2} < \binom{n}{3} p^3 (1-p)^{n-3}.

By solving this inequality, we see that the likelihood of winning changes with increasing weeks, often becoming counter-intuitive to initial assumptions.

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 \left( 1 - \frac{2}{n} \right) = \frac{2}{n(n-1)} - HSC - SSCE Mathematics Extension 1 - Question 4 - 2004 - Paper 1

Use mathematical induction to prove that for all integers n ≥ 3, $$ \left( 1 - \frac{2}{3} \right) \left( 1 - \frac{2}{4} \right) \left( 1 - \frac{2}{5} \right) \cdots \left( 1 - \frac{2}{n} \right) = \frac{2}{n(n-1)}. $$

Show that the tangents at the points P and Q meet at R, where R is the point (ap + aq, apq).

As P varies, the point Q is always chosen so that ∠POQ is a right angle, where O is the origin.

What is the probability that in the first 7 weeks Katie will win at least 1 prize?

Show that in the first 20 weeks Katie has a greater chance of winning exactly 2 prizes than of winning exactly 1 prize.

Most candidates assumed Katie participates in the prize drawing so that this week Katie has a greater chance of winning exactly 3 prizes than of winning exactly 2 prizes.

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