Math Problem Solver

\[ 1\frac{1}{7} + 1\frac{2}{7} + 2\frac{3}{7} = \frac{34}{7} = 4.\overline{857142} \]

The sum can be computed by rewriting each term over a common denominator, where the least common denominator of \(\frac{1}{1}\), \(\frac{1}{7}\), \(\frac{1}{1}\), \(\frac{2}{7}\), \(\frac{2}{1}\), and \(\frac{3}{7}\) is \(7\). \[ \begin{aligned} 1\frac{1}{7} + 1\frac{2}{7} + 2\frac{3}{7} &= \frac{1 \cdot 7}{1 \cdot 7} + \frac{1 \cdot 1}{7 \cdot 1} + \frac{1 \cdot 7}{1 \cdot 7} + \frac{2 \cdot 1}{7 \cdot 1} + \frac{2 \cdot 7}{1 \cdot 7} + \frac{3 \cdot 1}{7 \cdot 1} \\ &= \frac{7 + 1 + 7 + 2 + 14 + 3}{7} \\ &= \frac{34}{7} \end{aligned} \]