计算：$\frac {1}{15}+\frac {1}{105}+\frac {1}{315}+\frac {1}{693}+\frac {1}{1287}+\frac {1}{2145}$。
答案：$\frac{1}{15}+\frac{1}{105}+\frac{1}{315}+\frac{1}{693}+\frac{1}{1287}+\frac{1}{2145}=\frac{1}{1×3×5}+\frac{1}{3×5×7}+\frac{1}{5×7×9}+\cdots+\frac{1}{11×13×15}=\frac{1}{4}×(\frac{1}{1×3}-\frac{1}{3×5}+\frac{1}{3×5}-\frac{1}{5×7}+\frac{1}{5×7}-\frac{1}{7×9}+\cdots+\frac{1}{11×13}-\frac{1}{13×15})=\frac{1}{4}×(\frac{1}{1×3}-\frac{1}{13×15})=\frac{16}{195}$
解析：
$\frac{1}{15}+\frac{1}{105}+\frac{1}{315}+\frac{1}{693}+\frac{1}{1287}+\frac{1}{2145}$
$=\frac{1}{1×3×5}+\frac{1}{3×5×7}+\frac{1}{5×7×9}+\frac{1}{7×9×11}+\frac{1}{9×11×13}+\frac{1}{11×13×15}$
$=\frac{1}{4}×\left(\frac{1}{1×3}-\frac{1}{3×5}+\frac{1}{3×5}-\frac{1}{5×7}+\frac{1}{5×7}-\frac{1}{7×9}+\frac{1}{7×9}-\frac{1}{9×11}+\frac{1}{9×11}-\frac{1}{11×13}+\frac{1}{11×13}-\frac{1}{13×15}\right)$
$=\frac{1}{4}×\left(\frac{1}{1×3}-\frac{1}{13×15}\right)$
$=\frac{1}{4}×\left(\frac{1}{3}-\frac{1}{195}\right)$
$=\frac{1}{4}×\left(\frac{65}{195}-\frac{1}{195}\right)$
$=\frac{1}{4}×\frac{64}{195}$
$=\frac{16}{195}$
$=\frac{1}{1×3×5}+\frac{1}{3×5×7}+\frac{1}{5×7×9}+\frac{1}{7×9×11}+\frac{1}{9×11×13}+\frac{1}{11×13×15}$
$=\frac{1}{4}×\left(\frac{1}{1×3}-\frac{1}{3×5}+\frac{1}{3×5}-\frac{1}{5×7}+\frac{1}{5×7}-\frac{1}{7×9}+\frac{1}{7×9}-\frac{1}{9×11}+\frac{1}{9×11}-\frac{1}{11×13}+\frac{1}{11×13}-\frac{1}{13×15}\right)$
$=\frac{1}{4}×\left(\frac{1}{1×3}-\frac{1}{13×15}\right)$
$=\frac{1}{4}×\left(\frac{1}{3}-\frac{1}{195}\right)$
$=\frac{1}{4}×\left(\frac{65}{195}-\frac{1}{195}\right)$
$=\frac{1}{4}×\frac{64}{195}$
$=\frac{16}{195}$