Let `f (n) = sum _(r =2)^(n) (r )/(""^r C_(2) ""^(r+1) C_(2)) , a = lim _(x to oo) f (n) and x ^(2) -(2n -(1)/(2)) x +t =0` has two positive roots `alpha and beta.`
If value of `f (7) + f(8) is p /q` where p and q are relatively prime, then `(p-q)` is :

Options

A. 53
B. 55
C. 57
D. 59

Answer

To solve the problem step by step, we will evaluate the function \( f(n) \) and calculate \( f(7) + f(8) \).

### Step 1: Define the function \( f(n) \)
The function is defined as:
\[ f(n) = \sum_{r=2}^{n} \frac{r}{\binom{r}{2} \cdot \binom{r+1}{2}} \]

### Step 2: Simplify the terms in the summation
We know that:
\[ \binom{r}{2} = \frac{r(r-1)}{2} \quad \text{and} \quad \binom{r+1}{2} = \frac{(r+1)r}{2} \]
Thus, we can rewrite:
\[ \binom{r}{2} \cdot \binom{r+1}{2} = \frac{r(r-1)}{2} \cdot \frac{(r+1)r}{2} = \frac{r^2(r-1)(r+1)}{4} \]

### Step 3: Substitute back into the function
Now substituting back into the function:
\[ f(n) = \sum_{r=2}^{n} \frac{r}{\frac{r^2(r-1)(r+1)}{4}} = \sum_{r=2}^{n} \frac{4}{r(r-1)(r+1)} \]

### Step 4: Simplify the fraction
We can further simplify:
\[ \frac{4}{r(r-1)(r+1)} = \frac{4}{(r-1)r(r+1)} \]
This can be expressed using partial fractions:
\[ \frac{4}{(r-1)r(r+1)} = \frac{A}{r-1} + \frac{B}{r} + \frac{C}{r+1} \]

### Step 5: Find coefficients A, B, C
Multiplying through by the denominator \((r-1)r(r+1)\) and solving for \(A\), \(B\), and \(C\):
\[ 4 = A r(r+1) + B (r-1)(r+1) + C (r-1)r \]
Setting \(r = 1\), \(r = 0\), and \(r = -1\) gives us the values for \(A\), \(B\), and \(C\).

After solving, we find:
- \(A = 2\)
- \(B = -4\)
- \(C = 2\)

Thus,
\[ \frac{4}{(r-1)r(r+1)} = \frac{2}{r-1} - \frac{4}{r} + \frac{2}{r+1} \]

### Step 6: Evaluate \( f(n) \)
Now substituting back into the summation:
\[ f(n) = \sum_{r=2}^{n} \left( \frac{2}{r-1} - \frac{4}{r} + \frac{2}{r+1} \right) \]
This can be simplified using telescoping series.

### Step 7: Calculate \( f(7) \) and \( f(8) \)
Calculating \( f(7) \):
\[ f(7) = 2 \left( 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} \right) - 4 \left( 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} \right) + 2 \left( \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \right) \]

Calculating \( f(8) \):
\[ f(8) = 2 \left( 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} \right) - 4 \left( 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \right) + 2 \left( \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} \right) \]

### Step 8: Combine \( f(7) + f(8) \)
After calculating both \( f(7) \) and \( f(8) \), we combine them:
\[ f(7) + f(8) = \frac{p}{q} \]

### Step 9: Find \( p - q \)
Finally, we find \( p \) and \( q \) such that they are relatively prime, and compute \( p - q \).