Evaluate $\sum _{n=1}^{k}\frac{{\left(-4\right)}^{n+2}}{(n-2{)}^{2}}\prod _{i=n}^{2k}\frac{i}{i+1}$ for $k=3$ $\sum _{n=1}^{k}\frac{{\left(-4\right)}^{n+2}}{(n-2{)}^{2}}\prod _{i=n}^{2k}\frac{i}{i+1}$
How can I think this for k=3 ?
Here is my idea: $\prod _{i=n}^{6}\frac{i}{i+1}=\frac{1}{2}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}$
then
$\frac{3}{14}\sum _{n=1}^{3}\frac{{\left(-4\right)}^{n+2}}{{\left(n-2\right)}^{2}}=\frac{3}{14}.\left(\frac{{\left(-4\right)}^{3}}{{\left(1-2\right)}^{2}}+\frac{{\left(-4\right)}^{5}}{{\left(3-2\right)}^{2}}\right)=-233,1428571$
Is my idea correct?

*Note: * For $k=1$ to $3$ we have to evaulate the sum for $k\in\{1,2,3\}$:
\begin{align*}
\sum _{n=1}^{k}\frac{{\left(-4\right)}^{n+2}}{(n-2{)}^{2}}\prod _{i=n}^{2k}\frac{i}{i+1}\tag{1}
\end{align*}
Hint: Evaluation at $n=2$ is not feasible, since in that case we have division by zero, which is not admissible. In order to overcome this problem we can exclude the value which causes the indeterminate expressions;.

We consider the  valid expression:
\begin{align*}
\sum _{{n=1}\atop{n\ne 2}}^{k}\frac{{\left(-4\right)}^{n+2}}{(n-2{)}^{2}}
\prod _{i=n}^{2k}\frac{i}{i+1}\tag{2}
\end{align*}
which  can be simplified, since
\begin{align*}
\prod _{i=n}^{2k}\frac{i}{i+1}&=\frac{n}{n+1}\cdot\frac{n+1}{n+2}\cdots\frac{2k}{2k+1 }\\
&=\frac{n}{2k+2}
\end{align*}

