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 $n = 1$:

$LHS: 1 \times 2^{1} = 2$ $RHS: 2 + (1 - 1)2^{1 + 1} = 2 + 0 = 2$

So, the base case holds.

Inductive Step: Assume true for $n = k$:

$LHS: (1 \times 2^{k}) + (3 \times 2^{k}) + \cdots + (k \times 2^{k}) = 2 + (k - 1)2^{k + 1}$

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

$LHS: 2 + (k - 1)2^{k + 1} + ((k + 1) \times 2^{k + 1})$ $= 2 + k2^{k + 1}$ $= 2 + k2^{k + 2}$ $= 2 + (k + 1 - 1)2^{(k + 1) + 1}$

Thus, it holds for $k + 1$ and by induction, it is true for all integers $n \geq 1$.

Using the binomial probability formula, we can express the probability that exactly $k$ successes occur in $n$ independent Bernoulli trials:

$P(X = k) = C_{n}^{k} p^{k} (1 - p)^{n - k}$

For $n = 5$ (treadmills), $k = 3$, and $p = 0.65$:

$P(X = 3) = C_{5}^{3} (0.65)^{3} (0.35)^{2}$

Thus, the expression becomes:

$P(X = 3) = 10(0.65)^{3} (0.35)^{2}$

For this case, we combine both probabilities:

• The probability that exactly 3 treadmills are in use:

$C_{5}^{3} (0.65)^{3} (0.35)^{2}$

• The probability that no rowing machines are in use:

$C_{4}^{0} (0.4)^{0} (0.6)^{4} = (0.6)^{4}$

Combining these:

$P(X = 3, Y = 0) = C_{5}^{3} (0.65)^{3} (0.35)^{2} \cdot (0.6)^{4}$

From the binomial coefficient identities, we can combine:

$2022C_{80} + 2022C_{81} = 2022C_{82}$

This indicates:

$2022C_{80} + 2022C_{81} + 2022C_{193} = 2022C_{82} + 2022C_{193}$

This can be represented as:

$2022C_{80 + 81} + 2022C_{193} = 2022C_{82} + 2022C_{193} = 2022C_{q}$

Thus a possible solution is $p = 2022$ and $q = 81$.