`sum_(r=1)^10 r/(1-3r^2+r^4)`

Options

A. `-(50)/(109)`
B. `-(54)/(109)`
C. `-(55)/(111)`
D. `-(55)/(109)`

Answer

To solve the summation \( \sum_{r=1}^{10} \frac{r}{1 - 3r^2 + r^4} \), we will follow these steps:

### Step 1: Simplify the Denominator
First, we rewrite the denominator:
\[ 1 - 3r^2 + r^4 = r^4 - 3r^2 + 1 \]
We can rearrange this as:
\[ r^4 - 2r^2 + 1 - r^2 = (r^2 - 1)^2 - r^2 \]

### Step 2: Factor the Expression
Using the difference of squares, we factor the expression:
\[ \frac{r}{(r^2 - 1)^2 - r^2} = \frac{r}{(r^2 - 1 - r)(r^2 - 1 + r)} \]
This gives us:
\[ \frac{r}{(r^2 - r - 1)(r^2 + r - 1)} \]

### Step 3: Rewrite the Summation
Now, we can express the term \( \frac{r}{(r^2 - r - 1)(r^2 + r - 1)} \) in a form suitable for summation:
\[ \frac{r}{(r^2 - r - 1)(r^2 + r - 1)} = \frac{1}{2} \left( \frac{1}{r^2 - r - 1} - \frac{1}{r^2 + r - 1} \right) \]

### Step 4: Evaluate the Summation
Now, we can write the summation as:
\[ \sum_{r=1}^{10} \frac{1}{2} \left( \frac{1}{r^2 - r - 1} - \frac{1}{r^2 + r - 1} \right) \]
This can be simplified to:
\[ \frac{1}{2} \left( \sum_{r=1}^{10} \frac{1}{r^2 - r - 1} - \sum_{r=1}^{10} \frac{1}{r^2 + r - 1} \right) \]

### Step 5: Calculate Each Term
Now we calculate each term for \( r = 1 \) to \( 10 \):
1. For \( r = 1 \):
\[ T_1 = \frac{1}{2} \left( \frac{1}{1 - 1 - 1} - \frac{1}{1 + 1 - 1} \right) = \frac{1}{2} \left( \frac{1}{-1} - \frac{1}{1} \right) = \frac{1}{2} (-1 - 1) = -1 \]
2. For \( r = 2 \):
\[ T_2 = \frac{1}{2} \left( \frac{1}{4 - 2 - 1} - \frac{1}{4 + 2 - 1} \right) = \frac{1}{2} \left( \frac{1}{1} - \frac{1}{5} \right) = \frac{1}{2} \left( 1 - 0.2 \right) = \frac{0.8}{2} = 0.4 \]
3. Continue this for \( r = 3 \) to \( r = 10 \).

### Step 6: Sum All Terms
After calculating each term, we sum them up. Notice that many terms will cancel out due to the telescoping nature of the series.

### Final Result
After performing all calculations and summing the results, we find:
\[ \sum_{r=1}^{10} \frac{r}{1 - 3r^2 + r^4} = -\frac{55}{109} \]