Question

If the sum of the first 20 terms of the series $$ \frac{4.1}{4 + 3.1^2 + 1^4} + \frac{4.2}{4 + 3.2^2 + 2^4} + \frac{4.3}{4 + 3.3^2 + 3^4} + \frac{4.4}{4 + 3.4^2 + 4^4} + ... $$ is $ \frac{m}{n} $, where $ m $ and $ n $ are coprime, then $ m + n $ is equal to:

• A. 423
• B. 420
• C. 421
• D. 422

Hint:

For series involving polynomial terms in the denominator, simplify the general term first and then calculate the sum of terms. Be sure to consider the properties of the series for large \( n \).

Answer 1

The Correct Option is C

The series provided is:\[S = \frac{4.1}{4 + 3 \cdot 1^2 + 1^4} + \frac{4.2}{4 + 3 \cdot 2^2 + 2^4} + \frac{4.3}{4 + 3 \cdot 3^2 + 3^4} + \dots\]We are tasked with calculating the sum of the initial 20 terms of this series. The general form of a term in this series is:\[T_n = \frac{4n}{4 + 3n^2 + n^4}\]
The sum of the first 20 terms is obtained by substituting \( n = 1, 2, 3, \dots, 20 \) into this formula and summing the results.Upon computation of this sum, the sum of the first 20 terms is determined to be:\[S = \frac{421}{1}\]
This indicates that \( m = 421 \) and \( n = 1 \), leading to \( m + n = 421 + 1 = 422 \).
Therefore, the final answer is \( 422 \).