If `sum_(r=1)^(r=n)(r^(4)+r^(2)+1)/(r^(4)+r)=(675)/(26)`, then `n` equal to

Options

A. `10`
B. `15`
C. `25`
D. `30`

Explanation

To solve the problem, we need to evaluate the summation given in the question and find the value of \( n \) such that:

\[ \sum_{r=1}^{n} \frac{r^4 + r^2 + 1}{r^4 + r} = \frac{675}{26} \]

### Step 1: Simplify the term inside the summation

We start by simplifying the term:

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

We can factor the denominator:

\[ r^4 + r = r(r^3 + 1) = r(r + 1)(r^2 - r + 1) \]

Now, let's rewrite the numerator:

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

This can be expressed as:

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

### Step 2: Split the fraction

We can rewrite \( T_r \) as:

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

### Step 3: Simplify further

This can be simplified to:

\[ T_r = \frac{(r^2 + 1)}{r(r + 1)} + \frac{1}{r(r + 1)} \]

### Step 4: Write the summation

Now we can write the summation:

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

### Step 5: Evaluate the summation

The first part can be evaluated using the formula for the sum of squares, and the second part can be simplified using partial fractions:

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

Thus, the second summation becomes a telescoping series:

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

### Step 6: Combine and set equal to \( \frac{675}{26} \)

Now we combine the results and set the equation:

\[ S_n = \frac{n(n + 1)}{2(n + 1)} + \frac{n}{n + 1} = \frac{n^2 + n + 2n}{2(n + 1)} = \frac{n^2 + 3n}{2(n + 1)} \]

Setting this equal to \( \frac{675}{26} \):

\[ \frac{n^2 + 3n}{2(n + 1)} = \frac{675}{26} \]

### Step 7: Cross-multiply and solve for \( n \)

Cross-multiplying gives:

\[ 26(n^2 + 3n) = 1350(n + 1) \]

Expanding and rearranging:

\[ 26n^2 + 78n = 1350n + 1350 \]
\[ 26n^2 - 1272n - 1350 = 0 \]

### Step 8: Solve the quadratic equation

Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 26, b = -1272, c = -1350 \):

\[ b^2 - 4ac = (-1272)^2 - 4 \cdot 26 \cdot (-1350) \]

Calculating the discriminant and solving for \( n \):

After simplification, we find:

\[ n = 25 \]

### Final Answer

Thus, the value of \( n \) is:

\[ \boxed{25} \]