The sum of ` sum _(n=1)^(16) (n^(4))/(4n^(2) -1) ` is

Answer

To solve the problem of finding the sum

\[ \sum_{n=1}^{16} \frac{n^4}{4n^2 - 1} \]

we will follow these steps:

### Step 1: Simplify the Expression
We start with the expression:

\[ \frac{n^4}{4n^2 - 1} \]

We can factor the denominator:

\[ 4n^2 - 1 = (2n - 1)(2n + 1) \]

Thus, the expression becomes:

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

### Step 2: Rewrite the Fraction
Next, we can rewrite the fraction by multiplying and dividing by \(16\):

\[ \frac{n^4}{(2n - 1)(2n + 1)} = \frac{16n^4}{16(4n^2 - 1)} = \frac{16n^4}{(2n - 1)(2n + 1)} \]

### Step 3: Split the Fraction
Now we can split the fraction into two parts:

\[ \frac{16n^4 - 1 + 1}{(2n - 1)(2n + 1)} = \frac{16n^4 - 1}{(2n - 1)(2n + 1)} + \frac{1}{(2n - 1)(2n + 1)} \]

### Step 4: Apply the Difference of Squares
The first term can be simplified using the difference of squares:

\[ 16n^4 - 1 = (4n^2 - 1)(4n^2 + 1) \]

Thus, we have:

\[ \frac{(4n^2 - 1)(4n^2 + 1)}{(2n - 1)(2n + 1)} = \frac{(4n^2 - 1)}{(2n - 1)(2n + 1)}(4n^2 + 1) \]

### Step 5: Calculate the Summation
Now we can calculate the summation:

\[ \sum_{n=1}^{16} \left( \frac{(4n^2 + 1)}{(2n - 1)(2n + 1)} + \frac{1}{(2n - 1)(2n + 1)} \right) \]

Let’s denote:

- \( S_1 = \sum_{n=1}^{16} (4n^2 + 1) \)
- \( S_2 = \sum_{n=1}^{16} \frac{1}{(2n - 1)(2n + 1)} \)

### Step 6: Evaluate \( S_1 \)
Using the formula for the sum of squares:

\[ S_1 = 4 \sum_{n=1}^{16} n^2 + \sum_{n=1}^{16} 1 \]

We know:

\[ \sum_{n=1}^{N} n^2 = \frac{N(N + 1)(2N + 1)}{6} \]

For \( N = 16 \):

\[ \sum_{n=1}^{16} n^2 = \frac{16 \cdot 17 \cdot 33}{6} = 1496 \]

Thus,

\[ S_1 = 4 \cdot 1496 + 16 = 5984 + 16 = 6000 \]

### Step 7: Evaluate \( S_2 \)
For \( S_2 \):

\[ S_2 = \sum_{n=1}^{16} \frac{1}{(2n - 1)(2n + 1)} = \sum_{n=1}^{16} \left( \frac{1}{2n - 1} - \frac{1}{2n + 1} \right) \]

This is a telescoping series, and after simplification, we find that:

\[ S_2 = \frac{1}{1} - \frac{1}{33} = 1 - \frac{1}{33} = \frac{32}{33} \]

### Step 8: Combine Results
Now we combine \( S_1 \) and \( S_2 \):

\[ \sum_{n=1}^{16} \frac{n^4}{4n^2 - 1} = \frac{1}{16} (S_1 + S_2) = \frac{1}{16} (6000 + \frac{32}{33}) \]

### Step 9: Final Calculation
Calculating:

\[ = \frac{1}{16} \left( 6000 + \frac{32}{33} \right) = \frac{123376}{528} \text{ (after finding a common denominator)} \]

Thus, the final answer is:

\[ \frac{123376}{528} \]