Alan

Does this help:

 Remember that cos(90-α) is the same as sin(α)  (I've used α for angle HAB)

.

i^2 = -1

i^4 = 1

i^40 = (i^4)^10 = 1^10 = 1

(i = sqrt(-1))

.

It is 16910...0 where there are 135 zeros.

.

You could also use the double angle formula:

$$\tan(2\theta)=\frac{2\tan(\theta)}{1-\tan^2(\theta)}$$

If θ = 22.5° then 2θ = 45° and tan(45°) = 1 so you have:  

$$1=\frac{2\tan(22.5)}{1-\tan^2(22.5)}$$

This is a quadratic in tan(22.5°) the positive solution of which is √2 - 1 or 0.4142....

.

The angle HAB is given by tan^-1(CG/AG) so the horizontal (x) component of force is 2*sin(HAB) kN, and the vertical (y) component is -2*cos(HAB) kN  (assuming y is positive upwards).

Can you take it from there?

.

even x odd = even

.

"48 miles.    Mmmm Now that sounds like a stretch."

In reality there will be some air-drag resistance.  The resistive force tends to be proportional to velocityⁿ, and while the bullet is supersonic n can be as high as 6.  This would have a significant impact on the total distance travelled!

.

Arithmetic sequence is t(n) = a + (n-1)*d  where a is the first term, d is the difference and t(n) is the n'th term

So:

145 = a + 7d  from the 8'th term

607 = a + 49d from the 50'th term

Subtract the first equation from the second

462 = 42d

d = 462/42 = 11

Substitute this back into the first equation

145 = a + 7*11

a = 145 - 77 = 68

Hence t(129) = 68 + 128*11 = 1476

.

heureka has given the result for p equals any value, not necessarily between 0 and 1.  The sum isn't 1 when 0<p<1

The probabilistic expression (which is probably what was meant) is:

$$\sum_{k=0}^N(\frac{N}{k})p^{N-k}(1-p)^k$$

.