Use the principle of mathematical induction to show that for all integers $n \geq 1$, $$ 1 \times 2 + 2 \times 5 + 3 \times 8 + \cdots + n(3n-1) = n^2(n + 1). $$ When a particular biased coin is tossed, the probability of obtaining a head is $ \frac{3}{5} $. This coin is tossed 100 times. Let $X$ be the random variable representing the number of heads obtained. This random variable will have a binomial distribution. (i) Find the expected value, $E(X)$. (ii) By finding the variance, $Var(X)$, show that the standard deviation of $X$ is approximately 5. (iii) By using a normal approximation, find the approximate probability that $X$ is between 55 and 65. To complete a course, a student must choose and pass exactly three topics. There are eight topics from which to choose. Last year 400 students completed the course. Explain, using the pigeonhole principle, why at least eight students passed exactly the same three topics. Find $$ \int_0^{\pi/2} \cos 5x \sin 3x \,dx. $$ Find the curve which satisfies the differential equation $ \frac{dy}{dx} = -x + y $ and passes through the point (1, 0).

SSCE HSC Mathematics Extension 1 Trigonometric products as sums or differences: Use the principle of mathematica

Use the principle of mathematical induction to show that for all integers $n \geq 1$, $$ 1 \times 2 + 2 \times 5 + 3 \times 8 + \cdots + n(3n-1) = n^2(n + 1) - HSC - SSCE Mathematics Extension 1 - Question 12 - 2020 - Paper 1

Question 12

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Use the principle of mathematical induction to show that for all integers $n \geq 1$, $$ 1 \times 2 + 2 \times 5 + 3 \times 8 + \cdots + n(3n-1) = n^2(n + 1). $$ ... show full transcript

Worked Solution & Example Answer:Use the principle of mathematical induction to show that for all integers $n \geq 1$, $$ 1 \times 2 + 2 \times 5 + 3 \times 8 + \cdots + n(3n-1) = n^2(n + 1) - HSC - SSCE Mathematics Extension 1 - Question 12 - 2020 - Paper 1

Step 1

Use the principle of mathematical induction to show that for all integers $n \geq 1$, $$ 1 \times 2 + 2 \times 5 + 3 \times 8 + \cdots + n(3n-1) = n^2(n + 1). $$

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Answer

To prove by induction, we first check the base case where $n=1$ :

LHS: $1 \times 2 = 2$ RHS: $1^2(1 + 1) = 2$ .

Both sides are equal, hence the base case holds.

Now, assume it holds for $n=k$ :
$1 \times 2 + 2 \times 5 + \cdots + k(3k-1) = k^2(k + 1).$

We must show it holds for $n=k + 1$ :

LHS:

1 \times 2 + 2 \times 5 + \cdots + k(3k-1) + (k + 1)(3(k + 1)-1) =& k^2(k + 1) + (k + 1)(3k + 2)\\ =& (k + 1)(k + 3k + 2)\\ =& (k + 1)(k + 1)(k + 2). \end{align*}$$ Thus, $$LHS = (k + 1)^2((k + 1) + 1),$$ verifying our induction assumption.

Step 2

Find the expected value, $E(X)$.

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Answer

The expected value for a binomial distribution is given by: $E(X) = np,$
where $n = 100$ (number of trials) and $p = \frac{3}{5}$ .
Thus, $E(X) = 100 \times \frac{3}{5} = 60.$

Step 3

By finding the variance, $Var(X)$, show that the standard deviation of $X$ is approximately 5.

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Answer

The variance of a binomial distribution is given by: $Var(X) = np(1-p).$ Substituting, we get:

$Var(X) = 100 \times \frac{3}{5} \times \frac{2}{5} = 24.$
Therefore, the standard deviation is:

$\sigma = \sqrt{Var(X)} = \sqrt{24} \approx 4.899.$

Step 4

By using a normal approximation, find the approximate probability that $X$ is between 55 and 65.

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Answer

Using the normal approximation: $P(55 \leq X \leq 65) \approx P\left(\frac{55 - 60}{\sigma} < Z < \frac{65 - 60}{\sigma}\right) = P\left(-1.024 < Z < 1.024\right).$
Using the standard normal distribution, this yields: $\approx 0.668.$

Step 5

Explain, using the pigeonhole principle, why at least eight students passed exactly the same three topics.

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Answer

There are exactly $\binom{8}{3} = 56$ ways to choose 3 topics from 8. If 400 students completed the course, by the pigeonhole principle, at least: $\lceil \frac{400}{56} \rceil = 8$
students must have chosen the same combination of 3 topics.

Step 6

Find $$\int_0^{\pi/2} \cos 5x \sin 3x \,dx.$$

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Answer

Using the product-to-sum formulas:

$\cos A \sin B = \frac{1}{2} [\sin (A + B) - \sin (A - B)],$ we have:

$\int \cos 5x \sin 3x \,dx = \frac{1}{2} [\int \sin (8x) \,dx - \int \sin (2x) \,dx].$ Evaluating these integrals gives:

$= \frac{1}{2} \left[-\frac{1}{8} \cos 8x + \frac{1}{2} \cos 2x \right]_0^{\pi/2} = \frac{1}{2}\left[0 - \left(-\frac{1}{2}\right) \right] = \frac{1}{2}.$

Step 7

Find the curve which satisfies the differential equation $ \frac{dy}{dx} = -x + y $ and passes through the point (1, 0).

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Answer

Rearranging and separating variables gives: $\frac{dy}{y + x} = dx.$
Integrating both sides: $\ln |y + x| = x + C.$
Exponentiating yields: $y + x = ke^x.$
Using the condition that it passes through (1, 0): $0 + 1 = ke^1 \Rightarrow k = \frac{1}{e}.$
Thus the equation of the curve is: $y + x = \frac{1}{e}e^x \Rightarrow y = e^{x-1} - x.$

Use the principle of mathematical induction to show that for all integers $n \geq 1$, $$ 1 \times 2 + 2 \times 5 + 3 \times 8 + \cdots + n(3n-1) = n^2(n + 1) - HSC - SSCE Mathematics Extension 1 - Question 12 - 2020 - Paper 1

Worked Solution & Example Answer:Use the principle of mathematical induction to show that for all integers $n \geq 1$, $$ 1 \times 2 + 2 \times 5 + 3 \times 8 + \cdots + n(3n-1) = n^2(n + 1) - HSC - SSCE Mathematics Extension 1 - Question 12 - 2020 - Paper 1

Use the principle of mathematical induction to show that for all integers $n \geq 1$, $$ 1 \times 2 + 2 \times 5 + 3 \times 8 + \cdots + n(3n-1) = n^2(n + 1). $$

Find the expected value, $E(X)$.

By finding the variance, $Var(X)$, show that the standard deviation of $X$ is approximately 5.

By using a normal approximation, find the approximate probability that $X$ is between 55 and 65.

Explain, using the pigeonhole principle, why at least eight students passed exactly the same three topics.

Find $$\int_0^{\pi/2} \cos 5x \sin 3x \,dx.$$

Find the curve which satisfies the differential equation \( \frac{dy}{dx} = -x + y \) and passes through the point (1, 0).

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