Math Problem Solver

\[ \frac{1}{40} + \frac{9}{252} + \frac{4}{20} + \frac{4}{80} = \frac{87}{280} = 0.310\overline{714285} \]

The sum can be computed by rewriting each term over a common denominator, where the least common denominator of \(\frac{1}{40}\), \(\frac{9}{252}\), \(\frac{4}{20}\), and \(\frac{4}{80}\) is \(5040\). \[ \begin{aligned} \frac{1}{40} + \frac{9}{252} + \frac{4}{20} + \frac{4}{80} &= \frac{1 \cdot 126}{40 \cdot 126} + \frac{9 \cdot 20}{252 \cdot 20} + \frac{4 \cdot 252}{20 \cdot 252} + \frac{4 \cdot 63}{80 \cdot 63} \\ &= \frac{126 + 180 + 1008 + 252}{5040} \\ &= \frac{1566}{5040} \\&= \frac{87}{280} \end{aligned} \]