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

To show this, we utilize the Binomial theorem:

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

Adding these two expansions, we notice that odd powers of x cancel out, and we get:

$(1+x)^n + (1-x)^n = 2 \sum_{k \text{ even}} \binom{n}{k} x^k$

This can be simplified to:

$2[n + \binom{n}{2} + ... + \binom{n}{n}]$

To show this, we differentiate the previous result:

Taking the derivative of (1+x)^n and (1-x)^n with respect to x, we have:

$n(1+x)^{n-1} - n(1-x)^{n-1} = n[ (1+x)^{n-1} + (1-x)^{n-1} ]$

Consequently, using x = 1 or -1 will reveal the required identity.

Using the results from the previous parts, we substitute the expressions derived. By handling even terms appropriately using known properties of binomial coefficients, we conclude that:

$n(\binom{n}{2} + 3\binom{n}{4}) = n2^{n-3}.$

The horizontal range R can be derived by analyzing the motion:

Using the parametric equations: $R = V_{x} t = V\cos(θ) \cdot \frac{2V\sin(θ)}{g} = \frac{V^{2} \sin 2θ}{g}.$

Assuming V^{2} < 100g, substituting into our range formula gives:

$R = \frac{V^{2}\sin 2θ}{g} < 100$

This leads to an inequality which constrains θ to a value that results in R < 100 m.

Considering the maximum range occurs when θ = 45°, we check the values:

From the derived inequality, we evaluate: $\sin 2θ = 1$ and find the limiting angles, concluding with the required bounds.

The greatest height H occurs when the vertical component of velocity equals zero:

$H = \frac{V^{2}\sin^{2}(θ)}{2g}.$

As θ = 45° maximizes this, we substitute to find the maximum: $H_{max} = \frac{100}{2g}.$