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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

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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.

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    $\begingroup$ +1. Added an explanation at the end, if you do not like it please erase it. $\endgroup$ – Did Jan 10 '17 at 15:18
  • $\begingroup$ @Did: I like it. Thanks. $\endgroup$ – mathlove Jan 10 '17 at 15:20
  • $\begingroup$ @mathlove from where did the idea of rewriting $k+1$ come from? Some exercise you did? Inspiration? $\endgroup$ – RGS Jan 10 '17 at 16:07
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    $\begingroup$ @RSerrao: You can start from what you want to have. The form we want to have is something like $$k(k+1)(k+2)(k+1)=C(f(k)k(k+1)(k+2)-g(k)k(k+1)(k+2))$$where $f(k)=(k+3)(k+4)(k+5)\cdots, g(k)=(k-1)(k-2)(k-3)\cdots$ with a constant $C$. Dividing the both sides by $k(k+1)(k+2)$ gives $$k+1=C(f(k)-g(k))$$ Since the degree of the LHS is $1$, the degree of both $f(k)$ and $g(k)$ has to be $2$. So we are happy if we can have $$k+1=C((k+3)(k+4)-(k-1)(k-2))$$ which indeed holds for $C=1/10$. $\endgroup$ – mathlove Jan 11 '17 at 4:13
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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

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