2010-II-12

\begin{equation*}
\frac{0}{3},~\frac{-1}{4},~\frac{2}{5},~\frac{-3}{6},~\frac{4}{7}, \ldots
\end{equation*}

則可得

$\begin{array}{rcl}
T(1) & = & \dfrac{0}{3} \\
T(2) & = & \dfrac{-1}{4} \\
T(3) & = & \dfrac{2}{5} \\
T(4) & = & \dfrac{-3}{6} \\
T(5) & = & \dfrac{4}{7} \\
& \vdots &
\end{array}$

透過比較其項數及項內的數字，可得 $T(n)=(-1)^{n+1} \dfrac{n-1}{n+2}$。