Question

If the sum of the first 10 terms of the series $\frac{1}{1+1^4 \cdot 4} + \frac{2}{1+2^4 \cdot 4} + \frac{3}{1+3^4 \cdot 4} + \dots$ is $\frac{m}{n}$, $\text{gcd}(m, n) = 1$, then $m+n$ is equal to :

• A. 256
• B. 264
• C. 276
• D. 284

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This series involves a telescoping sum. The denominator $4r^4 + 1$ can be factorized using the Sophie Germain identity.
: Key Formula or Approach:
General term $T_r = \frac{r}{1 + 4r^4}$.
Identity: $4r^4 + 1 = (2r^2 + 1)^2 - (2r)^2 = (2r^2 - 2r + 1)(2r^2 + 2r + 1)$.
Step 2: Detailed Explanation:
Express $T_r$ as partial fractions:
$T_r = \frac{r}{(2r^2 - 2r + 1)(2r^2 + 2r + 1)} = \frac{1}{4} \left( \frac{(2r^2 + 2r + 1) - (2r^2 - 2r + 1)}{(2r^2 - 2r + 1)(2r^2 + 2r + 1)} \right)$
$T_r = \frac{1}{4} \left( \frac{1}{2r^2 - 2r + 1} - \frac{1}{2r^2 + 2r + 1} \right)$.
Let $f(r) = \frac{1}{2r^2 + 2r + 1}$. Then $f(r-1) = \frac{1}{2(r-1)^2 + 2(r-1) + 1} = \frac{1}{2r^2 - 2r + 1}$.
$S_{10} = \sum_{r=1}^{10} \frac{1}{4} [f(r-1) - f(r)] = \frac{1}{4} [f(0) - f(10)]$.
$f(0) = \frac{1}{1} = 1$.
$f(10) = \frac{1}{2(100) + 2(10) + 1} = \frac{1}{221}$.
$S_{10} = \frac{1}{4} [ 1 - \frac{1}{221} ] = \frac{1}{4} \cdot \frac{220}{221} = \frac{55}{221}$.
Here $m = 55, n = 221$. Since $221 = 13 \times 17$, $\text{gcd}(55, 221) = 1$.
$m + n = 55 + 221 = 276$.
Step 3: Final Answer:
The value of $m+n$ is 276.