1. Express the fraction $\frac{6}{n(n+1)(n+2)(n+3)}$ as the algebraic sum of four fractions with denominators $n$, $n+1$, $n+2$, and $n+3$, respectively. The result is $\qquad$.
\[
\begin{array}{l}
=1 \cdot \frac{1}{n}-\frac{3}{n+1}+\frac{3}{n+2}-\frac{1}{n+3} . \\ 
\frac{6}{n(n+1)(n+2)(n+3)} \\ 
=\frac{3}{n(n+3)}-\frac{3}{(n+1)(n+2)} \\ 
=\left(\frac{1}{n}-\frac{1}{n+3}\right)-\left(\frac{3}{n+1}-\frac{3}{n+2}\right) \\ 
=\frac{1}{n}-\frac{3}{n+1}+\frac{3}{n+2}-\frac{1}{n+3} .
\end{array}
\]