If the sum of the first 10 terms of the series frac4 cdot 11 + 4 cdot 1^4 + frac4 cdot 21 + 4 cdot 2^4 + frac4 cdot 31 + 4 cdot 3^4 + dots$\frac{4 \cdot 1}{1 + 4 \cdot 1^4} + \frac{4 \cdot 2}{1 + 4 \cdot 2^4} + \frac{4 \cdot 3}{1 + 4 \cdot 3^4} + \dots$ is fracmathrmmmathrmn$\frac{\mathrm{m}}{\mathrm{n}}$, where gcd(mathrmm, mathrmn) = 1$\gcd(\mathrm{m}, \mathrm{n}) = 1$, then mathrmm + mathrmn$\mathrm{m} + \mathrm{n}$ is equal to ____________.

### Related Formula textSophie Germain's algebraic factorization: 1 + 4r^4 = (2r^2 + 2r + 1)(2r^2 - 2r + 1)$$\text{Sophie Germain's algebraic factorization: } 1 + 4r^4 = (2r^2 + 2r + 1)(2r^2 - 2r + 1)$$ textTelescoping Series representation: T_r = f(r) - f(r+1)$$\text{Telescoping Series representation: } T_r = f(r) - f(r+1)$$ ### Core Logic This is a telescoping series sum. We expand the denominator using Sophie Germain's algebraic identity to write the general term as a difference of two consecutive rational expressions. ### Step 1: Write down the general term and factor The general term T_r$T_r$ of the series is: T_r = frac4r1 + 4r^4$$T_r = \frac{4r}{1 + 4r^4}$$ Using the factorization of 1+4r^4$1+4r^4$: T_r = frac4r(2r^2 - 2r + 1)(2r^2 + 2r + 1)$$T_r = \frac{4r}{(2r^2 - 2r + 1)(2r^2 + 2r + 1)}$$ Notice that the numerator 4r$4r$ is the exact difference of the two quadratic factors: (2r^2 + 2r + 1) - (2r^2 - 2r + 1) = 4r$$(2r^2 + 2r + 1) - (2r^2 - 2r + 1) = 4r$$ ### Step 2: Split the fraction into telescoping terms Rewrite the general term T_r$T_r$: T_r = frac(2r^2 + 2r + 1) - (2r^2 - 2r + 1)(2r^2 - 2r + 1)(2r^2 + 2r + 1)$$T_r = \frac{(2r^2 + 2r + 1) - (2r^2 - 2r + 1)}{(2r^2 - 2r + 1)(2r^2 + 2r + 1)}$$ T_r = frac12r^2 - 2r + 1 - frac12r^2 + 2r + 1 = f(r) - f(r+1)$$T_r = \frac{1}{2r^2 - 2r + 1} - \frac{1}{2r^2 + 2r + 1} = f(r) - f(r+1)$$ ### Step 3: Expand the sum and solve Sum the first 10 terms: - For r = 1$r = 1$: T_1 = frac11 - frac15$T_1 = \frac{1}{1} - \frac{1}{5}$ - For r = 2$r = 2$: T_2 = frac15 - frac113$T_2 = \frac{1}{5} - \frac{1}{13}$ - ... - For r = 10$r = 10$: T_10 = frac1181 - frac1221$T_{10} = \frac{1}{181} - \frac{1}{221}$ All intermediate terms cancel out: S_10 = 1 - frac1221 = frac220221 = fracmn$$S_{10} = 1 - \frac{1}{221} = \frac{220}{221} = \frac{m}{n}$$ Since 220$220$ and 221$221$ are coprime (their greatest common divisor is 1): m = 220 quad textand quad n = 221$$m = 220 \quad \text{and} \quad n = 221$$ m + n = 220 + 221 = 441$$m + n = 220 + 221 = 441$$ ### Pattern Recognition Sophie Germain identity: The expansion 4r^4 + 1 = (2r^2 - 2r + 1)(2r^2 + 2r + 1)$4r^4 + 1 = (2r^2 - 2r + 1)(2r^2 + 2r + 1)$ is highly common in telescoping series. Spotting this factorization collapses the sum instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series