By writing $\dfrac{2}{r(r^2-1)}$ in partial fractions, find an expression for $\displaystyle\sum_{r=2}^{n}\dfrac{2}{r(r^2-1)}$. A geometric series has first term $a$ and common ratio $r$, where $a$ and $r$ are non-zero and $r\ne 1$. The 3rd and 9th terms of the series are 448 and 7 respectively. Given also that the sum of the first $n$ terms is 1197, find the values of $a$, $r$ and $n$.

1. Express the given fraction in partial fractions

Factorise the denominator: $r(r^2-1) = r(r-1)(r+1)$.
Let the expression be:
$\frac{2}{r(r-1)(r+1)} = \frac{A}{r-1} + \frac{B}{r} + \frac{C}{r+1}$
Multiplying by $r(r-1)(r+1)$:
$2 = Ar(r+1) + B(r-1)(r+1) + Cr(r-1)$
Substitute $r = 1$:
$2 = A(1)(2) \implies A = 1$
Substitute $r = 0$:
$2 = B(-1)(1) \implies B = -2$
Substitute $r = -1$:
$2 = C(-1)(-2) \implies C = 1$
Hence,
$\frac{2}{r(r^2-1)} = \frac{1}{r-1} - \frac{2}{r} + \frac{1}{r+1}$

2. List terms of the series to observe the telescoping pattern

$\sum_{r=2}^{n} \left( \frac{1}{r-1} - \frac{2}{r} + \frac{1}{r+1} \right)$
Write out the first few and last few terms:
When $r=2$: $\quad \frac{1}{1} - \frac{2}{2} + \frac{1}{3}$
When $r=3$: $\quad \frac{1}{2} - \frac{2}{3} + \frac{1}{4}$
When $r=4$: $\quad \frac{1}{3} - \frac{2}{4} + \frac{1}{5}$
$\dots$
When $r=n-1$: $\quad \frac{1}{n-2} - \frac{2}{n-1} + \frac{1}{n}$
When $r=n$: $\quad \frac{1}{n-1} - \frac{2}{n} + \frac{1}{n+1}$

3. Cancel intermediate terms to find the sum

Notice that for any $k$ from $3$ to $n-1$, the fraction with denominator $k$ appears as $\frac{1}{k}$ (from $r=k-1$), $-\frac{2}{k}$ (from $r=k$), and $\frac{1}{k}$ (from $r=k+1$), summing to $0$.
The uncancelled terms are:

• From $r=2$: $\frac{1}{1} - \frac{2}{2}$
• From $r=3$: $\frac{1}{2}$
• From $r=n-1$: $\frac{1}{n}$
• From $r=n$: $-\frac{2}{n} + \frac{1}{n+1}$

Summing these up gives:
$1 - 1 + \frac{1}{2} + \frac{1}{n} - \frac{2}{n} + \frac{1}{n+1}$
$= \frac{1}{2} - \frac{1}{n} + \frac{1}{n+1}$

4. Combine the result into a single fraction

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

1. Formulate equations for the 3rd and 9th terms of the geometric series.

The $n$th term of a geometric series is $u_n = ar^{n-1}$.
$ar^2 = 448 \quad \text{--- (1)}$
$ar^8 = 7 \quad \text{--- (2)}$

2. Solve the equations simultaneously to find the possible values of $r$ and the value of $a$.

Dividing equation (2) by equation (1):
$\frac{ar^8}{ar^2} = \frac{7}{448}$
$r^6 = \frac{1}{64}$
$r = \frac{1}{2} \text{ or } r = -\frac{1}{2}$
Substitute $r^2 = \frac{1}{4}$ into equation (1):
$a\left(\frac{1}{4}\right) = 448 \implies a = 1792$

3. Test the first case where $r = \frac{1}{2}$ to see if it yields a valid integer $n$.

Using the sum of the first $n$ terms formula $S_n = \frac{a(1-r^n)}{1-r}$:
$\frac{1792\left(1 - \left(\frac{1}{2}\right)^n\right)}{1 - \frac{1}{2}} = 1197$
$3584\left(1 - \left(\frac{1}{2}\right)^n\right) = 1197$
$\left(\frac{1}{2}\right)^n = 1 - \frac{1197}{3584} = \frac{341}{512}$
Since $n$ must be an integer and $\frac{341}{512}$ is not a perfect power of $\frac{1}{2}$, $r = \frac{1}{2}$ is rejected.

4. Solve for $n$ using the valid ratio $r = -\frac{1}{2}$.

$\frac{1792\left(1 - \left(-\frac{1}{2}\right)^n\right)}{1 - \left(-\frac{1}{2}\right)} = 1197$
$\frac{1792}{\frac{3}{2}}\left(1 - \left(-\frac{1}{2}\right)^n\right) = 1197$
$1 - \left(-\frac{1}{2}\right)^n = 1197 \times \frac{3}{3584} = \frac{513}{512}$
$\left(-\frac{1}{2}\right)^n = 1 - \frac{513}{512} = -\frac{1}{512}$
Since $-\frac{1}{512} = \left(-\frac{1}{2}\right)^9$, we have $n = 9$.