If `a_(n)=(-2)/(4n^(2)-16n+15)`,then `a_(1)+a_(2)+......+a_(25)` is equal to :

Options

A. `(52)/(147)`
B. `(50)/(141)`
C. `(51)/(144)`
D. `(49)/(138)`

Answer

To solve the problem, we start with the given sequence:

\[ a_n = \frac{-2}{4n^2 - 16n + 15} \]

We need to find the sum:

\[ S = a_1 + a_2 + \ldots + a_{25} = \sum_{n=1}^{25} a_n \]

### Step 1: Factor the denominator
First, we factor the quadratic expression in the denominator:

\[ 4n^2 - 16n + 15 = (2n - 5)(2n - 3) \]

### Step 2: Rewrite the term
Now we can rewrite \(a_n\):

\[ a_n = \frac{-2}{(2n - 5)(2n - 3)} \]

### Step 3: Partial fraction decomposition
Next, we perform partial fraction decomposition:

\[ \frac{-2}{(2n - 5)(2n - 3)} = \frac{A}{2n - 5} + \frac{B}{2n - 3} \]

Multiplying through by the denominator \((2n - 5)(2n - 3)\) gives:

\[ -2 = A(2n - 3) + B(2n - 5) \]

### Step 4: Solve for A and B
Expanding the right side:

\[ -2 = (2A + 2B)n + (-3A - 5B) \]

Setting up the system of equations by equating coefficients:

1. \(2A + 2B = 0\)
2. \(-3A - 5B = -2\)

From the first equation, we have:

\[ A + B = 0 \implies B = -A \]

Substituting \(B = -A\) into the second equation:

\[ -3A - 5(-A) = -2 \implies -3A + 5A = -2 \implies 2A = -2 \implies A = -1 \]

Thus, \(B = 1\). Therefore, we can write:

\[ a_n = \frac{-1}{2n - 5} + \frac{1}{2n - 3} \]

### Step 5: Rewrite the sum
Now we can express the sum \(S\):

\[ S = \sum_{n=1}^{25} a_n = \sum_{n=1}^{25} \left( \frac{-1}{2n - 5} + \frac{1}{2n - 3} \right) \]

### Step 6: Split the sum
This can be split into two separate sums:

\[ S = -\sum_{n=1}^{25} \frac{1}{2n - 5} + \sum_{n=1}^{25} \frac{1}{2n - 3} \]

### Step 7: Evaluate the sums
Now we evaluate each sum:

1. The first sum:

\[ -\sum_{n=1}^{25} \frac{1}{2n - 5} = -\left( \frac{1}{-3} + \frac{1}{-1} + \frac{1}{1} + \ldots + \frac{1}{45} \right) \]

2. The second sum:

\[ \sum_{n=1}^{25} \frac{1}{2n - 3} = \left( \frac{1}{-1} + \frac{1}{1} + \frac{1}{3} + \ldots + \frac{1}{47} \right) \]

### Step 8: Combine the results
After evaluating these sums, we find that many terms will cancel out. The remaining terms will yield:

\[ S = \frac{1}{3} - \frac{1}{47} + \text{(cancellation of intermediate terms)} \]

### Final Result
After simplifying the remaining terms, we find:

\[ S = \frac{50}{147} \]

### Conclusion
Thus, the final answer is:

\[ \boxed{\frac{50}{147}} \]