The sum of the series, \(\frac{1}{2.3} \cdot 2+\frac{2}{3.4} \cdot 2^{2}+\frac{3}{4.5} \cdot 2^{3}+\ldots\) upto \(n\) terms is

Given series is \(\frac{1}{2 \cdot 3} \cdot 2^{1}+\frac{2}{3.4} \cdot 2^{2}+\frac{3}{4.5} \cdot 2^{3}+\ldots\) upto \(n\) terms The \(n\)th term of the series is
\(T_{n}=\frac{n}{(n+1)(n+2)} \cdot 2^{n}=\left\{\frac{2 \cdot 2^{n}}{(n+2)}-\frac{2^{n}}{(n+1)}\right\}\)
On putting \(n=1,2,3 \ldots\), we get
\(T_{1}=2 \cdot \frac{2^{1}}{3}-\frac{2^{1}}{2}, T_{2}=\frac{2 \cdot 2^{2}}{4}-\frac{2^{2}}{3}, T_{3}=\frac{2 \cdot 2^{3}}{5}-\frac{2^{3}}{4}\)
On adding, we get
\(\begin{aligned}
&\quad S_{n}=T_{1}+T_{2}+T_{3}+\ldots+T_{n} \\
&\Rightarrow \quad S_{n}=\frac{2 \cdot 2^{n}}{(n+2)}-\frac{2}{2}=\frac{2^{n+1}}{(n+2)}-1
\end{aligned}\)