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 is $T_r = \frac{r}{1 + 4r^4}$. This denominator can be factored using Sophie Germain's Identity: $1 + 4r^4 = (1 + 2r^2 + 2r)(1 + 2r^2 - 2r)$. This allows us to use the method of differences (telescoping series).

Step 2: Key Formula or Approach:
1. $T_r = \frac{r}{(2r^2 + 2r + 1)(2r^2 - 2r + 1)}$. 2. Express $T_r$ as a difference: $T_r = \frac{1}{4} \left[ \frac{1}{2r^2 - 2r + 1} - \frac{1}{2r^2 + 2r + 1} \right]$.

Step 3: Detailed Explanation:
1. Let $f(r) = \frac{1}{2r^2 - 2r + 1}$. Then $f(r+1) = \frac{1}{2(r+1)^2 - 2(r+1) + 1} = \frac{1}{2r^2 + 2r + 1}$. 2. The sum $S_{10} = \sum_{r=1}^{10} \frac{1}{4} [f(r) - f(r+1)]$. 3. By telescoping: $S_{10} = \frac{1}{4} [f(1) - f(11)]$. 4. Calculate values: - $f(1) = \frac{1}{2(1)^2 - 2(1) + 1} = 1$. - $f(11) = \frac{1}{2(11)^2 - 2(11) + 1} = \frac{1}{242 - 22 + 1} = \frac{1}{221}$. 5. $S_{10} = \frac{1}{4} [1 - \frac{1}{221}] = \frac{1}{4} \cdot \frac{220}{221} = \frac{55}{221}$. 6. $m = 55, n = 221$. Sum $m+n = 55 + 221 = 276$.

Step 4: Final Answer:
The value of $(m+n)$ is 276.