If `T_( r )=( r )/(r^(4)+4)` and `S_(n)=underset(r=1)overset(n)sum t_( r )`. Then the value of `37S_(5)-(7)/(26)` is equal to

Options

A. 13
B. `25//26`
C. `1//2`
D. `7//36`

Answer

To solve the problem, we need to find the value of \( 37S_5 - \frac{7}{26} \), where \( S_n = \sum_{r=1}^{n} T_r \) and \( T_r = \frac{r}{r^4 + 4} \).

### Step 1: Simplify \( T_r \)

We start with the expression for \( T_r \):

\[ T_r = \frac{r}{r^4 + 4} \]

We can rewrite \( r^4 + 4 \) as \( r^4 + 4r^2 + 4 - 4r^2 \):

\[ T_r = \frac{r}{(r^2 + 2)^2 - (2r)^2} \]

This can be factored using the difference of squares:

\[ T_r = \frac{r}{(r^2 + 2 - 2r)(r^2 + 2 + 2r)} = \frac{r}{(r^2 - 2r + 2)(r^2 + 2r + 2)} \]

### Step 2: Partial Fraction Decomposition

We can express \( T_r \) in terms of partial fractions:

\[ T_r = \frac{A}{r^2 - 2r + 2} + \frac{B}{r^2 + 2r + 2} \]

Multiplying through by the denominator \( (r^2 - 2r + 2)(r^2 + 2r + 2) \) gives:

\[ r = A(r^2 + 2r + 2) + B(r^2 - 2r + 2) \]

Expanding and collecting like terms, we can find \( A \) and \( B \).

### Step 3: Solve for \( A \) and \( B \)

By substituting suitable values for \( r \) or by equating coefficients, we can find:

1. Set \( r = 0 \) to find \( A + B = 0 \).
2. Set \( r = 1 \) to find \( A(1 + 2 + 2) + B(1 - 2 + 2) = 1 \).
3. Solve the resulting equations to find \( A \) and \( B \).

After solving, we find:

\[ A = \frac{1}{4}, \quad B = -\frac{1}{4} \]

Thus,

\[ T_r = \frac{1/4}{r^2 - 2r + 2} - \frac{1/4}{r^2 + 2r + 2} \]

### Step 4: Calculate \( S_5 \)

Now we can calculate \( S_5 \):

\[ S_5 = \sum_{r=1}^{5} T_r = \frac{1}{4} \sum_{r=1}^{5} \left( \frac{1}{r^2 - 2r + 2} - \frac{1}{r^2 + 2r + 2} \right) \]

Calculating each term:

- For \( r = 1 \): \( T_1 = \frac{1}{4} \left( \frac{1}{1} - \frac{1}{5} \right) = \frac{1}{4} \cdot \frac{4}{5} = \frac{1}{5} \)
- For \( r = 2 \): \( T_2 = \frac{1}{4} \left( \frac{1}{2} - \frac{1}{6} \right) = \frac{1}{4} \cdot \frac{1}{3} = \frac{1}{12} \)
- For \( r = 3 \): \( T_3 = \frac{1}{4} \left( \frac{1}{4} - \frac{1}{8} \right) = \frac{1}{4} \cdot \frac{1}{8} = \frac{1}{32} \)
- For \( r = 4 \): \( T_4 = \frac{1}{4} \left( \frac{1}{6} - \frac{1}{10} \right) = \frac{1}{4} \cdot \frac{1}{15} = \frac{1}{60} \)
- For \( r = 5 \): \( T_5 = \frac{1}{4} \left( \frac{1}{8} - \frac{1}{12} \right) = \frac{1}{4} \cdot \frac{1}{24} = \frac{1}{96} \)

Now summing these values:

\[ S_5 = \frac{1}{5} + \frac{1}{12} + \frac{1}{32} + \frac{1}{60} + \frac{1}{96} \]

### Step 5: Calculate \( 37S_5 - \frac{7}{26} \)

After calculating \( S_5 \), we multiply by 37 and subtract \( \frac{7}{26} \):

\[ 37S_5 - \frac{7}{26} \]

Finding a common denominator and simplifying will yield the final answer.