Evaluate $$\int_{3}^{4} (x + 2) \sqrt{3 - x} \, dx$$ using the substitution $u = x - 3$ - HSC - SSCE Mathematics Extension 1 - Question 12 - 2023 - Paper 1

1. Base Case: For $n = 1$, the left-hand side equals $1 \times 2^1 = 2$ and the right-hand side equals $2 + (1 - 1)2^{1 + 1} = 2$. Both sides are equal.

2. Inductive Step: Assume it is true for $n = k$: $\sum_{j=1}^{k} (j \times 2^j) = 2 + (k - 1)2^{k + 1}$ We need to show for $n = k + 1$: $\sum_{j=1}^{k + 1} (j \times 2^j) = 2 + (k)(2^{k + 2})$

3. Adding $(k + 1) \times 2^{k + 1}$ to both sides of the assumption: $\sum_{j=1}^{k + 1} (j \times 2^j) = (2 + (k - 1)2^{k + 1}) + (k + 1)2^{k + 1}$ Which simplifies to: $= 2 + k2^{k + 1} + 2^{k + 1} = 2 + k2^{k + 2}$

Thus, the statement holds for $n = k + 1$, completing the induction.

The probability can be modeled using the binomial distribution:

Let $p = 0.65$ (probability treadmill is in use), $n = 5$ (total treadmills), and $k = 3$ (treadmills in use).

The expression is:

$P(X = 3) = {5 \choose 3} (0.65)^3 (0.35)^{2}$ where ${n \choose k}$ is the binomial coefficient.

Using the same approach:

1. The probability at least 3 treadmills in use: $P(X = 3) = {5 \choose 3} (0.65)^3 (0.35)^{2}$

2. No rowing machines in use: Probability is given by $(0.4)^{0} (0.6)^4$

Thus, the combined probability: $P(X = 3) * P(Y = 0) = {5 \choose 3} (0.65)^3 (0.35)^{2} * (0.6)^4$

Applying the properties of binomial coefficients:

1. Use the identities: $2022 C_{30} + 2022 C_{31} = 2022 C_{31} + 2022 C_{32} = 2022 C_{32}$

This implies: $2022 C_{30} + 2022 C_{31} = 2022 C_{80}$

From the problem statement, we can equate: $P = 2022$ and find suitable values of $p$ and $q$ easily, for example, take $p = 81$ and $q = 1933$ to satisfy the equation.