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

Using the formula for the expected value in a binomial distribution:

$E(X) = n \cdot p$

d where $n = 100$ (number of tosses) and $p = \frac{3}{5}$ (probability of heads):

$E(X) = 100 \cdot \frac{3}{5} = 60.$

The variance for a binomial distribution is given by:

$Var(X) = n \cdot p \cdot (1 - p).$

Substituting values:

$Var(X) = 100 \cdot \frac{3}{5} \cdot \left(1 - \frac{3}{5}\right) = 100 \cdot \frac{3}{5} \cdot \frac{2}{5} = 24.$

Thus, the standard deviation, $\sigma$, is:

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

Using the normal approximation:

Convert to z-scores:

For $X = 55$: $z_1 = \frac{55 - 60}{\sigma} = \frac{-5}{5} = -1.$
For $X = 65$: $z_2 = \frac{65 - 60}{\sigma} = \frac{5}{5} = 1.$

Using the standard normal distribution table, $P(-1 < Z < 1) \approx 0.6827,$ which means the approximate probability that $X$ is between 55 and 65 is $68.27\%$.

Considering the number of ways to choose 3 topics from 8 topics:

$\binom{8}{3} = \frac{8!}{3!(8 - 3)!} = 56.$

Since last year 400 students completed the course, by the pigeonhole principle, if there are 400 students and 56 combinations of topics, at least one combination must have more than one student:

$\frac{400}{56} \approx 7.14.$

Thus, at least 8 students must have chosen the same combination of topics.

Using the product-to-sum identities:

$\int \cos A \sin B \ dx = \frac{1}{2}\int(\sin(A + B) - \sin(A - B)) \ dx.$

Applying: $\int \cos 5x \sin 3x \ dx = \frac{1}{2}\left[\int \sin(8x) \ dx - \int \sin(2x) \ dx\right].$

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

To solve the differential equation:

Rearranging gives: $\frac{dy}{dx} - y = -x.$

Using integrating factors:

Let $e^{\int -1 dx} = e^{-x}$:

Multiplying through by this integrating factor, we have: $e^{-x} \frac{dy}{dx} - e^{-x}y = -xe^{-x}.$

Now integrating yields: $e^{-x}y = \int -xe^{-x} dx.$

By parts: $\int -xe^{-x} dx = -xe^{-x} - \int -e^{-x} dx = -xe^{-x} + e^{-x} + C.$

Thus, $y = (1 - x + C)e^{x}$. Using the passed point $(1, 0)$ to find $C$: $$0 = (1 - 1 + C)e^{1} \Rightarrow C = 0.$

Finally, the solution is:
$y = (1 - x)e^{x},$ which is the equation of the curve.