The sum of series $1\cdot 3\cdot 2^2+2\cdot 4\cdot 3^2+3\cdot 5\cdot 4^2+\cdots \cdots n$ terms The sum of series $1\cdot 3\cdot 2^2+2\cdot 4\cdot 3^2+3\cdot 5\cdot 4^2+\cdots \cdots n$ terms 
i have calculate $a_{k} = k(k+2)(k+1)^2$
so $\displaystyle \sum^{n}_{k=1}a_{k} = \sum^{n}_{k=1}k(k+1)^2(k+2)$
i wan,t be able go further, could some help me with this, thanks

Hint:
Observe that
\begin{align*}
a_k&=k(k+2)(k+1)^2\\
&=(k+1-1)(k+1+1)(k+1)^2\\
&=\left[(k+1)^2-1\right](k+1)^2\\
&=(k+1)^4-(k+1)^2
\end{align*}
Now, here is something that can be useful in order to compute the sum of the fourth powers and squares 
Geometric interpretation for sum of fourth powers
