A particle is moving along the x-axis in simple harmonic motion centred at the origin - HSC - SSCE Mathematics Extension 1 - Question 13 - 2017 - Paper 1

Using the Binomial Theorem, we can express both terms as follows:

$(1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k$ $(1 - x)^n = \sum_{k=0}^{n} \binom{n}{k} (-x)^k$

By adding these equations together, we find that terms where k is odd cancel out, leading to:

$(1 + x)^n + (1 - x)^n = 2 \sum_{k=0}^{\frac{n}{2}} \binom{n}{2k} x^{2k} = 2\left[n \binom{n}{0} + \binom{n}{2} x^2 + ... + \binom{n}{n} x^n\right].$

Using the result from part (i):

Substituting the binomial expansions directly into:

$n[(1 + x)^n - (1 - x)^n] = n \left( 2 \sum_{k=0}^{\frac{n}{2}} \binom{n}{2k+1} x^{2k+1} \right)$

which eventually leads us to: $= 2 \left[n 2^{n - 1} \binom{n}{4} + \binom{n}{2} x^2 + ... + \binom{n}{n} x^n\right].$

To show this inequality, we first simplify it:

Multiply through by 12: $6n \leq 5n$ This means that n is positive and thus the inequality holds for all positive even integers.

The horizontal range R of a projectile launched at an angle θ is given by:

$R = \frac{V^2 \sin 2\theta}{g}$

This is derived from substituting the equations of motion for horizontal distance, considering the time of flight.

Given that: $R = \frac{V^2 \sin 2\theta}{g} < 100$ Substituting the condition leads us to: $V^2 < 100g \implies R < 100\text{m}$ This verifies that a speed less than the threshold results in a range less than 100 m.

Using the range formula: If V^2 = 200g, then substituting back into the formula gives: \sin 2θ = \frac{R g}{V^2}, which gives us the bounds on θ indicating: $\frac{π}{12} \leq θ \leq \frac{5π}{12}$.

Maximum height reaches when: $H = \frac{V^2 \sin^2 \theta}{2g}$ Plugging in values and maximizing with respect to θ will yield the greatest height. Setting up the derivative and solving gives the optimal angle.