Given that $\displaystyle\sum_{r=1}^{n} \frac{1}{r(r+1)} = \frac{n}{n+1}$.

1. Express the required sum in a form that matches the given identity by changing the index.

Let $k = r - 1$. Then $r = k + 1$.
When $r = 2$, $k = 1$.
When $r = n$, $k = n - 1$.
$\sum_{r=2}^{n} \frac{1}{r(r-1)} = \sum_{k=1}^{n-1} \frac{1}{(k+1)k} = \sum_{k=1}^{n-1} \frac{1}{k(k+1)}$

2. Apply the given identity and simplify to obtain the result.

Using the given identity $\displaystyle\sum_{r=1}^{n} \frac{1}{r(r+1)} = \frac{n}{n+1}$ by replacing $n$ with $n-1$:
$\sum_{k=1}^{n-1} \frac{1}{k(k+1)} = \frac{n-1}{(n-1)+1}$
$= \frac{n-1}{n}$
$= \frac{n}{n} - \frac{1}{n}$
$= 1 - \frac{1}{n} \text{ (Shown)}$

1. Separate the sum into two individual summations and substitute the result from part (a).

$\sum_{r=2}^{n} \left( \frac{1}{2^r} + \frac{1}{r(r-1)} \right) = \sum_{r=2}^{n} \frac{1}{2^r} + \sum_{r=2}^{n} \frac{1}{r(r-1)}$
Using the result from part (a):
$\sum_{r=2}^{n} \frac{1}{r(r-1)} = 1 - \frac{1}{n}$

2. Evaluate the first summation as a geometric series.

The series $\sum_{r=2}^{n} \frac{1}{2^r}$ is a finite geometric progression.

• First term, $a = \frac{1}{2^2} = \frac{1}{4}$
• Common ratio, $r = \frac{1}{2}$
• Number of terms, $N = n - 2 + 1 = n - 1$

Applying the sum formula $S_N = \frac{a(1 - r^N)}{1 - r}$:
$\sum_{r=2}^{n} \frac{1}{2^r} = \frac{\frac{1}{4} \left(1 - \left(\frac{1}{2}\right)^{n-1}\right)}{1 - \frac{1}{2}}$
$= \frac{\frac{1}{4} \left(1 - \left(\frac{1}{2}\right)^{n-1}\right)}{\frac{1}{2}}$
$= \frac{1}{2} \left( 1 - \left(\frac{1}{2}\right)^{n-1} \right)$
$= \frac{1}{2} - \frac{1}{2} \left(\frac{1}{2}\right)^{n-1} = \frac{1}{2} - \left(\frac{1}{2}\right)^n$

3. Combine the evaluated parts and simplify.

$\sum_{r=2}^{n} \left( \frac{1}{2^r} + \frac{1}{r(r-1)} \right) = \left( \frac{1}{2} - \left(\frac{1}{2}\right)^n \right) + \left( 1 - \frac{1}{n} \right)$
$= \frac{3}{2} - \left(\frac{1}{2}\right)^n - \frac{1}{n}$