Question

If the sum of the first 10 terms of the series \( \frac{1}{1 + 4 \cdot 1^4} + \frac{2}{1 + 4 \cdot 2^4} + \frac{3}{1 + 4 \cdot 3^4} + \dots \) is \( \frac{m}{n} \) (where m, n are coprime), then (m + n) is:

• A. 264
• B. 276
• C. 284
• D. 256

Hint:

Sophie Germain's Identity: $a^4 + 4b^4 = (a^2 + 2b^2 + 2ab)(a^2 + 2b^2 - 2ab)$. It is a common trick in competitive exam series problems.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The general term of the series is \(T_r = \frac{r}{1 + 4r^4}\).
We need to factorize the denominator to simplify the expression, often leading to a telescopic series.
Step 2: Key Formula or Approach:
Using the identity \(4r^4 + 1 = (2r^2 + 1)^2 - (2r)^2 = (2r^2 + 2r + 1)(2r^2 - 2r + 1)\).
Now, express the numerator in terms of the factors of the denominator:
\[ T_r = \frac{r}{(2r^2 + 2r + 1)(2r^2 - 2r + 1)} \]
\[ T_r = \frac{1}{4} \left[ \frac{(2r^2 + 2r + 1) - (2r^2 - 2r + 1)}{(2r^2 + 2r + 1)(2r^2 - 2r + 1)} \right] \]
\[ T_r = \frac{1}{4} \left[ \frac{1}{2r^2 - 2r + 1} - \frac{1}{2r^2 + 2r + 1} \right] \]
Step 3: Detailed Explanation:
The sum \(S_{10} = \sum_{r=1}^{10} T_r\):
\[ S_{10} = \frac{1}{4} \left[ \left(\frac{1}{1} - \frac{1}{5}\right) + \left(\frac{1}{5} - \frac{1}{13}\right) + \dots + \left(\frac{1}{2(10)^2 - 2(10) + 1} - \frac{1}{2(10)^2 + 2(10) + 1}\right) \right] \]
This is a telescopic series where terms cancel out:
\[ S_{10} = \frac{1}{4} \left[ 1 - \frac{1}{221} \right] \]
\[ S_{10} = \frac{1}{4} \cdot \frac{220}{221} = \frac{55}{221} \]
Here \(m = 55\) and \(n = 221\).
Checking if they are coprime: \(55 = 5 \times 11\), \(221 = 13 \times 17\). They share no common factors.
Sum \(m + n = 55 + 221 = 276\).
Step 4: Final Answer:
The value of \(m + n\) is 276.