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

Base case: For $n = 1$: $LHS = 1 \times 2^{1} = 2$ $RHS = 2 + (1 - 1)2^{2} = 2$ Thus, $LHS = RHS$.

Inductive step: Assume true for $n = k$: $LHS = (1 \times 2^{2}) + (2 \times 2^{2}) + \ldots + (k \times 2^{k}) = 2 + (k - 1)2^{k + 1}$ Now for $n = k + 1$: $LHS = (1 \times 2^{2}) + (2 \times 2^{2}) + \ldots + (k \times 2^{k}) + ((k+1) \times 2^{k+1})$ Substituting from the assumption: $= 2 + (k - 1)2^{k + 1} + (k + 1)2^{k + 1}$ This simplifies to: $= 2 + k 2^{k + 1} = 2 + (k+1 - 1)2^{k + 2}$ Hence, proven for $n = k + 1$. Therefore, it holds for all integers $n \geq 1$.

Given that each treadmill is used 65% of the time, the probability for exactly 3 out of 5 treadmills can be expressed using the binomial formula: $P(X = k) = C(n, k) p^k (1 - p)^{n - k}$ Here, $n = 5, k = 3$, and $p = 0.65$: $P(X = 3) = C(5, 3)(0.65)^3(0.35)^2$

Using the same binomial formula, the probability of having exactly 3 treadmills in use, alongside having no rowing machines (which have a usage probability of 0.40), can be expressed as: $P(X = 3, Y = 0) = C(5, 3)(0.65)^3(0.35)^2 \cdot C(4, 0)(0.40)^0(0.60)^4$ Simplifying gives: $= C(5, 3)(0.65)^3(0.35)^2 \cdot 1 \cdot (0.60)^4$

From the given identity, we can deduce: $2022C_{80} + 2022C_{81} = 2022C_{81} + 2022C_{80} + 2022C_{1934}$ This means: $P = 2022\ and \ q = 81$ Thus, a possible solution is $p = 2022$, $q = 81$.