Sum to n terms the series whose nth terms is `(1)/(4n^(2)-1)`

Answer

To find the sum to n terms of the series whose nth term is given by \( A_n = \frac{1}{4n^2 - 1} \), we can follow these steps:

### Step 1: Rewrite the nth term
The nth term can be factored as:
\[ A_n = \frac{1}{4n^2 - 1} = \frac{1}{(2n - 1)(2n + 1)} \]

### Step 2: Use Partial Fraction Decomposition
We can express \( A_n \) in terms of partial fractions:
\[ A_n = \frac{1}{(2n - 1)(2n + 1)} = \frac{A}{2n - 1} + \frac{B}{2n + 1} \]
To find \( A \) and \( B \), we multiply through by the denominator \( (2n - 1)(2n + 1) \):
\[ 1 = A(2n + 1) + B(2n - 1) \]

### Step 3: Solve for A and B
Expanding the right-hand side gives:
\[ 1 = (2A + 2B)n + (A - B) \]
Setting coefficients equal, we have:
1. \( 2A + 2B = 0 \) (coefficient of \( n \))
2. \( A - B = 1 \) (constant term)

From the first equation, \( A + B = 0 \) implies \( B = -A \). Substituting into the second equation:
\[ A - (-A) = 1 \implies 2A = 1 \implies A = \frac{1}{2}, \quad B = -\frac{1}{2} \]

### Step 4: Rewrite \( A_n \) with the values of A and B
Thus, we can rewrite \( A_n \) as:
\[ A_n = \frac{1/2}{2n - 1} - \frac{1/2}{2n + 1} \]

### Step 5: Sum the series
Now, we can find the sum of the first n terms:
\[ S_n = \sum_{k=1}^{n} A_k = \sum_{k=1}^{n} \left( \frac{1/2}{2k - 1} - \frac{1/2}{2k + 1} \right) \]
This is a telescoping series. When we write out the first few terms, we see:
\[ S_n = \frac{1}{2} \left( \left( \frac{1}{1} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{5} \right) + \left( \frac{1}{5} - \frac{1}{7} \right) + \ldots + \left( \frac{1}{2n-1} - \frac{1}{2n+1} \right) \right) \]

### Step 6: Simplify the sum
Most terms will cancel out, leaving us with:
\[ S_n = \frac{1}{2} \left( 1 - \frac{1}{2n + 1} \right) \]
Thus, we can simplify:
\[ S_n = \frac{1}{2} - \frac{1}{2(2n + 1)} = \frac{1}{2} - \frac{1}{4n + 2} \]

### Final Answer
The sum of the first n terms is:
\[ S_n = \frac{n}{2(2n + 1)} \]