If `I(r)=r(r^2-1)`, then `sum_(r=2)^n 1/(I(r))` is equal to

Options

A. `1/4 (1-1/(n(n+1)))`
B. `1/4 (2-1/(n(n+1)))`
C. `1/4 (1-2/(n(n+1)))`
D. `1/4 (2+ 1/(n(n+1)))`

Answer

To solve the problem, we need to evaluate the sum:

\[ \sum_{r=2}^{n} \frac{1}{I(r)} \]

where \( I(r) = r(r^2 - 1) \).

### Step 1: Substitute \( I(r) \)

First, we substitute the expression for \( I(r) \):

\[ I(r) = r(r^2 - 1) = r(r-1)(r+1) \]

Thus, we can rewrite the sum as:

\[ \sum_{r=2}^{n} \frac{1}{I(r)} = \sum_{r=2}^{n} \frac{1}{r(r^2 - 1)} = \sum_{r=2}^{n} \frac{1}{r(r-1)(r+1)} \]

### Step 2: Use Partial Fraction Decomposition

Next, we can use partial fraction decomposition to simplify the term:

\[ \frac{1}{r(r-1)(r+1)} = \frac{A}{r-1} + \frac{B}{r} + \frac{C}{r+1} \]

Multiplying through by the denominator \( r(r-1)(r+1) \) gives:

\[ 1 = A r(r+1) + B (r-1)(r+1) + C r(r-1) \]

### Step 3: Solve for Coefficients

To find the coefficients \( A, B, C \), we can choose convenient values for \( r \):

1. Let \( r = 1 \):
\[ 1 = A(1)(2) + B(0) + C(0) \implies A = \frac{1}{2} \]

2. Let \( r = 0 \):
\[ 1 = A(0) + B(-1)(1) + C(0) \implies -B = 1 \implies B = -1 \]

3. Let \( r = -1 \):
\[ 1 = A(-2)(0) + B(0) + C(-1)(-2) \implies 2C = 1 \implies C = \frac{1}{2} \]

Thus, we have:

\[ \frac{1}{r(r-1)(r+1)} = \frac{1/2}{r-1} - \frac{1}{r} + \frac{1/2}{r+1} \]

### Step 4: Substitute Back into the Sum

Now we can substitute back into the sum:

\[ \sum_{r=2}^{n} \left( \frac{1/2}{r-1} - \frac{1}{r} + \frac{1/2}{r+1} \right) \]

### Step 5: Split the Sum

This can be split into three separate sums:

\[ \frac{1}{2} \sum_{r=2}^{n} \frac{1}{r-1} - \sum_{r=2}^{n} \frac{1}{r} + \frac{1}{2} \sum_{r=2}^{n} \frac{1}{r+1} \]

### Step 6: Evaluate Each Sum

1. The first sum \( \sum_{r=2}^{n} \frac{1}{r-1} \) is:

\[ \sum_{r=1}^{n-1} \frac{1}{r} \]

2. The second sum \( \sum_{r=2}^{n} \frac{1}{r} \) is:

\[ \sum_{r=2}^{n} \frac{1}{r} \]

3. The third sum \( \sum_{r=2}^{n} \frac{1}{r+1} \) is:

\[ \sum_{r=3}^{n+1} \frac{1}{r} \]

### Step 7: Combine the Results

Putting it all together, we have:

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

### Step 8: Simplify

After simplification, we can observe that many terms will cancel out. The final result will be:

\[ \frac{1}{4} \left( 1 - \frac{2}{n(n+1)} \right) \]

### Final Result

Thus, the value of the sum is:

\[ \frac{1}{4} \left( 1 - \frac{2}{n(n+1)} \right) \]