Example 2 Note
$$
\begin{array}{l}
S=\frac{2}{1 \times 3}+\frac{2^{2}}{3 \times 5}+\cdots+\frac{2^{49}}{97 \times 99}, \\
T=\frac{1}{3}+\frac{2}{5}+\frac{2^{2}}{7}+\cdots+\frac{2^{48}}{99} .
\end{array}
$$

Then the value of $S-T$ is ( ).
(A) $\frac{2^{49}}{99}$
(B) $1-\frac{2^{49}}{99}$
(C) $\frac{2^{49}}{99}-1$
(D) $\frac{2^{49}}{99}+1$
Solve: By observation and induction, we get the general term
$$
f(n)=\frac{2^{n}}{(2 n-1)(2 n+1)}-\frac{2^{n-1}}{2 n+1} \text {. }
$$

Decomposing the general term, we get
$$
\begin{array}{l}
f(n)=2^{n-1} \cdot \frac{(2 n+1)-(2 n-1)}{(2 n-1)(2 n+1)}-\frac{2^{n-1}}{2 n+1} \\
=2^{n-1}\left(\frac{1}{2 n-1}-\frac{1}{2 n+1}\right)-\frac{2^{n-1}}{2 n+1} \\
=\frac{2^{n-1}}{2 n-1}-\frac{2^{n}}{2 n+1} .
\end{array}
$$

Substituting $n=1,2, \cdots, 49$ respectively and adding them up, we get
$S-T$
$$
\begin{array}{l}
=\left(\frac{1}{1}-\frac{2}{3}\right)+\left(\frac{2}{3}-\frac{2^{2}}{5}\right)+\cdots+\left(\frac{2^{48}}{97}-\frac{2^{99}}{99}\right) \\
=1-\frac{2^{49}}{99} .
\end{array}
$$

Therefore, the answer is B.