Prove by induction the statement a + ar + ... + arⁿ−1 = a((1−rⁿ)/(1-r)) : r != 1. (b) (i) Express 1/(r(r+1)) - 1/((r+1)(r+2)) in the form A/(r(r+1)(r+2)). (ii) Hence show that sum from r=1 to n of 1/(r(r+1)(r+2)) = 1/4 - 1/(2(n+1)(n+2)).

The common denominator is $r(r+1)(r+2)$.

$$\frac{1}{r(r+1)} - \frac{1}{(r+1)(r+2)} = \frac{1 \cdot (r+2)}{r(r+1)(r+2)} - \frac{1 \cdot r}{r(r+1)(r+2)}$$

$$= \frac{(r+2) - r}{r(r+1)(r+2)}$$ $$= \frac{2}{r(r+1)(r+2)}$$

Comparing this to $\frac{A}{r(r+1)(r+2)}$, we find $A=2$.
Thus, $\frac{1}{r(r+1)} - \frac{1}{(r+1)(r+2)} = \boxed{\frac{2}{r(r+1)(r+2)}}$.

b)(ii) Hence show that $\sum \frac{1}{r(r+1)(r+2)} = \frac{1}{4} - \frac{1}{2(n+1)(n+2)}$.

From part (b)(i), we have $\frac{2}{r(r+1)(r+2)} = \frac{1}{r(r+1)} - \frac{1}{(r+1)(r+2)}$.
Dividing both sides by 2, we get the general term of the series:

$$\frac{1}{r(r+1)(r+2)} = \frac{1}{2} \left( \frac{1}{r(r+1)} - \frac{1}{(r+1)(r+2)} \right)$$

The sum is $\sum_{r=1}^{n} \frac{1}{2} \left( \frac{1}{r(r+1)} - \frac{1}{(r+1)(r+2)} \right)$.

$$S_n = \frac{1}{2} \sum_{r=1}^{n} \left( \frac{1}{r(r+1)} - \frac{1}{(r+1)(r+2)} \right)$$

Expanding the sum:

$$S_n = \frac{1}{2} \left[ \left( \frac{1}{1(2)} - \frac{1}{2(3)} \right) + \left( \frac{1}{2(3)} - \frac{1}{3(4)} \right) + \dots + \left( \frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)} \right) \right]$$

All intermediate terms cancel out, leaving only the first and last terms:

$$S_n = \frac{1}{2} \left[ \frac{1}{1(2)} - \frac{1}{(n+1)(n+2)} \right]$$ $$S_n = \frac{1}{2} \left[ \frac{1}{2} - \frac{1}{(n+1)(n+2)} \right]$$

$$S_n = \frac{1}{4} - \frac{1}{2(n+1)(n+2)}$$

This matches the statement to be shown.
Therefore, $\sum_{r=1}^{n} \frac{1}{r(r+1)(r+2)} = \boxed{\frac{1}{4} - \frac{1}{2(n+1)(n+2)}}$.