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 :
Since we have
$$(k+3)(k+4)-(k-2)(k-1)=10k+10$$
$$\implies \frac{(k+3)(k+4)-(k-2)(k-1)}{10}=k+1$$
we get
$$\begin{align}k(k+2)(k+1)^2&=k(k+1)(k+2)\color{red}{(k+1)}\\\\&=k(k+1)(k+2)\cdot\color{red}{\frac{1}{10}((k+3)(k+4)-(k-2)(k-1))}\\\\&=\frac{1}{10}(k(k+1)(k+2)(k+3)(k+4)-(k-2)(k-1)k(k+1)(k+2))\\\\&=\frac1{10}(a_{k+4}-a_{k+2})\end{align}$$ where $$a_k=k(k-1)(k-2)(k-3)(k-4)$$
Now simplify your sum using the idea of telescoping series.
