Question

If the sum of the first 10 terms of the series $$ \frac{4.1}{1 + 4.1^4} + \frac{4.2}{1 + 4.2^4} + \frac{4.3}{1 + 4.3^4} + \cdots $$ is $ \frac{m}{n} $, where $ \gcd(m, n) = 1 $, then $ m + n $ is equal to ........

Hint:

In such problems, ensure to break down the series terms clearly and check each term’s calculation. This helps in summing the terms and eventually determining \( m + n \).

Answer 1

Correct Answer: 76

The general term \( T_r \) of the series is given by: \[ T_r = \frac{4r}{1 + 4r^4} \] Hence, for the first few terms: \[ T_1 = \frac{4}{1 + 4 \times 1^4} = \frac{4}{5} \] \[ T_2 = \frac{4}{(2^2 + 2 \times 2 + 1)} = \frac{4}{13} \] \[ T_3 = \frac{4}{(3^2 + 3 \times 2 + 1)} = \frac{4}{21} \] \[ \vdots \] Now, the sum of the first 10 terms is given by: \[ S_{10} = T_1 + T_2 + \cdots + T_{10} \] Thus, for \( m \) and \( n \), we find: \[ S_{10} = \frac{1}{181} + \frac{220}{221} = \frac{220}{221} \] Hence, \( m + n = 220 + 221 = 441 \).