Question

The value of the sum of r / (4 + r⁴) for r = 1 to 10 is:

• A. \( \dfrac{4832}{123565} \)
• B. \( \dfrac{4565}{12322} \)
• C. \( \dfrac{4565}{12321} \)
• D. \( \dfrac{4564}{12321} \)

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The general term of the series has a denominator \( r^4 + 4 \), which is a classic form that can be factorized using Sophie Germain's Identity. Once factorized, we use the method of differences (telescoping series) to find the sum.
Step 2: Key Formula or Approach:
1. Sophie Germain's Identity: \( r^4 + 4 = (r^2 + 2)^2 - (2r)^2 = (r^2 + 2r + 2)(r^2 - 2r + 2) \).
2. Express the general term \( T_r \) as a difference: \( T_r = \frac{1}{4} [ \frac{1}{r^2 - 2r + 2} - \frac{1}{r^2 + 2r + 2} ] \).
Step 3: Detailed Explanation:
1. Let \( f(r) = r^2 + 2r + 2 \). Then \( f(r-2) = (r-2)^2 + 2(r-2) + 2 = r^2 - 4r + 4 + 2r - 4 + 2 = r^2 - 2r + 2 \). 2. The general term becomes: \[ T_r = \frac{r}{(r^2 + 2r + 2)(r^2 - 2r + 2)} = \frac{1}{4} \left[ \frac{(r^2 + 2r + 2) - (r^2 - 2r + 2)}{(r^2 + 2r + 2)(r^2 - 2r + 2)} \right] \] \[ T_r = \frac{1}{4} \left[ \frac{1}{r^2 - 2r + 2} - \frac{1}{r^2 + 2r + 2} \right] \] 3. Summing from \( r=1 \) to \( 10 \): \[ S_{10} = \frac{1}{4} \sum_{r=1}^{10} \left( \frac{1}{r^2 - 2r + 2} - \frac{1}{r^2 + 2r + 2} \right) \] \[ S_{10} = \frac{1}{4} \left[ \left( \frac{1}{1} - \frac{1}{5} \right) + \left( \frac{1}{2} - \frac{1}{10} \right) + \dots + \left( \frac{1}{82} - \frac{1}{122} \right) \right] \] Wait, let's re-examine the terms: \( r^2 - 2r + 2 = (r-1)^2 + 1 \) and \( r^2 + 2r + 2 = (r+1)^2 + 1 \). \[ T_r = \frac{1}{4} \left[ \frac{1}{(r-1)^2 + 1} - \frac{1}{(r+1)^2 + 1} \right] \] 4. Expanding the sum: \[ S_{10} = \frac{1}{4} \left[ \left( \frac{1}{0^2+1} - \frac{1}{2^2+1} \right) + \left( \frac{1}{1^2+1} - \frac{1}{3^2+1} \right) + \dots + \left( \frac{1}{9^2+1} - \frac{1}{11^2+1} \right) \right] \] \[ S_{10} = \frac{1}{4} \left[ \left( 1 + \frac{1}{2} \right) - \left( \frac{1}{10^2+1} + \frac{1}{11^2+1} \right) \right] \] \[ S_{10} = \frac{1}{4} \left[ \frac{3}{2} - \frac{1}{101} - \frac{1}{122} \right] = \frac{4565}{12321} \]
Step 4: Final Answer:
The sum is \( \frac{4565}{12321} \).