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 $$\int_0^{\frac{\pi}{2}} \cos 5x \sin 3x \, dx.$$ 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 Integrals involving trigonometric substitution: Use the principle of mathematica

Use the principle of mathematical induction to show that for all integers $n \geq 1$, $$1 \times 2 + 2 \times 5 + 3 \times 8 + \cdots + n(3n-1) = n^2(n + 1)$$ When a particular biased coin is tossed, the probability of obtaining a head is $\frac{3}{5}$ - 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 $n \geq 1$, $$1 \times 2 + 2 \times 5 + 3 \times 8 + \cdots + n(3n-1) = n^2(n + 1)$$ When... show full transcript

Worked Solution & Example Answer:Use the principle of mathematical induction to show that for all integers $n \geq 1$, $$1 \times 2 + 2 \times 5 + 3 \times 8 + \cdots + n(3n-1) = n^2(n + 1)$$ When a particular biased coin is tossed, the probability of obtaining a head is $\frac{3}{5}$ - HSC - SSCE Mathematics Extension 1 - Question 12 - 2020 - Paper 1

Step 1

Use the principle of mathematical induction to show that for all integers $n \geq 1$

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Answer

To prove the statement, we need to first verify the base case for $n = 1$ :

yielding $1 \times 2 = 1^2(1 + 1) = 2$ . The base case is true.

Now assume the statement is true for some $n = k$ : $1 \times 2 + 2 \times 5 + 3 \times 8 + \cdots + k(3k - 1) = k^2(k + 1)$ .

Next, we need to show it holds for $n = k + 1$ : $1 \times 2 + 2 \times 5 + 3 \times 8 + \cdots + k(3k - 1) + (k + 1)(3(k + 1) - 1) = k^2(k + 1) + (k + 1)(3k + 2).$ After simplification: $= (k + 1)(k + 2)$ , which is equal to $(k + 1)^2((k + 1) + 1)$ .

By induction, the formula holds for all integers $n \geq 1$ .

Step 2

Find the expected value, $E(X)$

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Answer

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

Step 3

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

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The variance for a binomial distribution is given by: $Var(X) = np(1 - p).$
Substituting the values: $Var(X) = 100 \times \frac{3}{5} \times \frac{2}{5} = 24.$
Thus, the standard deviation is: $\sigma = \sqrt{Var(X)} = \sqrt{24} \approx 5.$

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: We convert to standard normal: $Z = \frac{X - E(X)}{\sigma}.$
Calculating for $X = 55$ and $X = 65$ gives: For $X = 55$ ,
$Z_{55} = \frac{55 - 60}{5} = -1.$
For $X = 65$ ,
$Z_{65} = \frac{65 - 60}{5} = 1.$
Thus, we evaluate: $P(55 < X < 65) = P(-1 < Z < 1) \approx 68\%.$

Step 5

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

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There are $inom{8}{3} = 56$ different combinations of three topics that a student can choose. With 400 students and only 56 combinations, by the pigeonhole principle, at least one combination must have at least $rac{400}{56} \approx 7.14$ students. Thus, at least eight students must have passed the same three topics.

Step 6

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

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Using the product-to-sum identities, we have: $\cos A \sin B = \frac{1}{2}(\sin(A + B) - \sin(A - B)).$
Thus, $\int_0^{\frac{\pi}{2}} \cos 5x \sin 3x \, dx = \frac{1}{2} \left[ \int_0^{\frac{\pi}{2}} \sin(8x) - \int_0^{\frac{\pi}{2}} \sin(2x) \right].$
Calculating leads to: $$= \frac{1}{2} \left[-\frac{1}{8}(\sin(8x)) - \frac{1}{2}(\sin(2x)) \right]_0^{\frac{\pi}{2}} = \frac{1}{2} \left[0 + 0 - 0 - 0 \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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Separating variables: $\frac{dy}{dx} = y - x \\ => \frac{dy}{y - x} = dx.$
Integrating both sides gives: $\ln |y - x| = x+C.$
Exponentiating gives: $y - x = Ce^x.$
Evaluating with initial condition (1, 0): $0 - 1 = Ce^1 \Rightarrow C = -\frac{1}{e}.$
Thus, the final curve is: $y = -\frac{1}{e}e^x + x.$

Use the principle of mathematical induction to show that for all integers $n \geq 1$

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 $\int_0^{\frac{\pi}{2}} \cos 5x \sin 3x \, dx$

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

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