$ \begin {aligned}&\frac {3}{4}+\frac {3}{28}+\frac {3}{70}+\frac {3}{130}+\frac {3}{208}\\=&1-\frac {1}{4}+\frac {1}{4}-\frac {1}{7}+\frac {1}{7}-\frac {1}{10}+\frac {1}{10}-\frac {1}{13}+\frac {1}{13}-\frac {1}{16}\\=&1-\frac {1}{16}\\=&\frac {15}{16}\end {aligned}$