It is given that $\sum_{r=1}^n \dfrac{2}{r(r+1)(r+2)} = \dfrac{1}{2} - \dfrac{1}{(n+1)(n+2)}.$ State the sum to infinity of the series. Find the sum of the first $n$ terms of the series $\dfrac{2}{17\times18\times19} + \dfrac{2}{18\times19\times20} + \dfrac{2}{19\times20\times21} + \cdots.$ Find $\sum_{r=2}^n \dfrac{1}{r(r^2-1)}.$ The sequence $u_1,u_2,u_3,\dots$ is defined by $u_1 = 3$, $u_{n+1} = 2 - \dfrac{4}{u_n}$, $n \ge 1$. Find the values of $u_2$, $u_3$, $u_4$ and $u_{100}$. Hence find $\sum_{r=1}^{3n-2} u_r$ in terms of $n$, simplifying your answer.

1. Evaluate the limit of the sum to $n$ terms as $n \to \infty$

As $n \to \infty$, $(n+1)(n+2) \to \infty$, so $\dfrac{1}{(n+1)(n+2)} \to 0$.

$\text{Sum to infinity} = \lim_{n \to \infty} \left( \dfrac{1}{2} - \dfrac{1}{(n+1)(n+2)} \right) = \dfrac{1}{2}$

1. Express the required sum in sigma notation as the difference of two sums starting from $r=1$.

Let $S_n$ be the sum of the first $n$ terms. The first term starts at $r=17$, and the $n$-th term will be at $r = n+16$.

$S_n = \sum_{r=17}^{n+16} \dfrac{2}{r(r+1)(r+2)}$
$S_n = \sum_{r=1}^{n+16} \dfrac{2}{r(r+1)(r+2)} - \sum_{r=1}^{16} \dfrac{2}{r(r+1)(r+2)}$

2. Apply the given sum formula to evaluate both parts and simplify.

Substitute $n = n+16$ and $n = 16$ into the given formula $\sum_{r=1}^n \dfrac{2}{r(r+1)(r+2)} = \dfrac{1}{2} - \dfrac{1}{(n+1)(n+2)}$:

$S_n = \left( \dfrac{1}{2} - \dfrac{1}{(n+16+1)(n+16+2)} \right) - \left( \dfrac{1}{2} - \dfrac{1}{(16+1)(16+2)} \right)$
$S_n = \left( \dfrac{1}{2} - \dfrac{1}{(n+17)(n+18)} \right) - \left( \dfrac{1}{2} - \dfrac{1}{17 \times 18} \right)$
$S_n = \dfrac{1}{17 \times 18} - \dfrac{1}{(n+17)(n+18)}$
$S_n = \dfrac{1}{306} - \dfrac{1}{(n+17)(n+18)}$

1. Factorise the denominator to match the structure of the given series.

$\sum_{r=2}^n \dfrac{1}{r(r^2-1)} = \sum_{r=2}^n \dfrac{1}{(r-1)r(r+1)}$

2. Perform an index shift by letting $k = r-1$ to align with the given identity.

$\sum_{r=2}^n \dfrac{1}{(r-1)r(r+1)} = \sum_{k=1}^{n-1} \dfrac{1}{k(k+1)(k+2)}$

3. Apply the given sum formula, adjusting for the constant multiplier and the new upper limit.

$\begin{aligned} \sum_{k=1}^{n-1} \dfrac{1}{k(k+1)(k+2)} &= \dfrac{1}{2} \sum_{k=1}^{n-1} \dfrac{2}{k(k+1)(k+2)} \\ &= \dfrac{1}{2} \left[ \dfrac{1}{2} - \dfrac{1}{((n-1)+1)((n-1)+2)} \right] \\ &= \dfrac{1}{4} - \dfrac{1}{2n(n+1)} \end{aligned}$

1. Calculate the values of $u_2$, $u_3$, and $u_4$ using the given recurrence relation.

$u_2 = 2 - \dfrac{4}{u_1} = 2 - \dfrac{4}{3} = \dfrac{2}{3}$

$u_3 = 2 - \dfrac{4}{u_2} = 2 - \dfrac{4}{\frac{2}{3}} = 2 - 6 = -4$

$u_4 = 2 - \dfrac{4}{u_3} = 2 - \dfrac{4}{-4} = 2 + 1 = 3$

2. Observe the pattern to deduce $u_{100}$.

Since $u_4 = u_1 = 3$, the sequence is periodic and repeats every 3 terms ($u_{n+3} = u_n$).
To find $u_{100}$, we determine the position in the 3-term cycle:
$100 = 33 \times 3 + 1$
Therefore, $u_{100} = u_1 = 3$.

1. Determine the sum of one complete cycle of the periodic sequence.

From part (b)(i), the sequence is periodic with a period of 3, where $u_1 = 3$, $u_2 = \frac{2}{3}$, and $u_3 = -4$.

The sum of one complete cycle of 3 terms is:
$u_1 + u_2 + u_3 = 3 + \frac{2}{3} - 4 = -\frac{1}{3}$

2. Express the total sum in terms of full cycles and the remaining term, then simplify.

The series has $3n-2$ terms. We can write this as $3(n-1) + 1$, which means the sum consists of $n-1$ complete cycles of 3 terms, plus 1 additional term. Due to the periodicity, this final term corresponds to the start of a new cycle, which is $u_1$.

$\begin{aligned} \sum_{r=1}^{3n-2} u_r &= (n-1)\left(-\frac{1}{3}\right) + u_1 \\ &= -\frac{1}{3}(n-1) + 3 \\ &= -\frac{1}{3}n + \frac{1}{3} + 3 \\ &= \frac{10 - n}{3} \end{aligned}$