Colored Sum

Regard $A=1 \cdot 2 \cdot 3 \cdot 4 + 2 \cdot 3 \cdot 4 \cdot 5 + \ldots + n \cdot (n+1) \cdot (n+2) \cdot (n+3)$ $\iff 5A=1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 + 2 \cdot 3 \cdot 4 \cdot 5 \cdot 5 + \ldots + n \cdot (n+1) \cdot (n+2) \cdot (n+3) \cdot 5$ $\iff 5A=1 \cdot 2 \cdot 3 \cdot 4 \cdot 5+2\cdot 3 \cdot 4 \cdot 5 \cdot (6-1)~+~ \ldots~+~ (n-1) \cdot n \cdot (n+1) \cdot (n+2) \cdot [(n+3)-(n-2)]~+~n \cdot (n+1) \cdot (n+2) \cdot (n+3) \cdot [(n+4)-(n-1)]$ $\iff 5A=\color{#D61F06}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5} \color{#3D99F6}{+2 \cdot 3 \cdot 4 \cdot 5 \cdot 6}\color{#D61F06}{-2\cdot 3 \cdot 4 \cdot 5 \cdot 1}+ \ldots \color{#69047E}{+(n-1) \cdot n\cdot(n+1)\cdot(n+2)\cdot(n+3)} \color{#20A900}{-(n-1)\cdot n\cdot(n+1)\cdot(n+2)\cdot(n-2)} \color{#69047E}\color{#CEBB00}{+n\cdot(n+1)\cdot(n+2)\cdot(n+3)\cdot(n+4)} \color{#69047E}{-n \cdot (n+1) \cdot (n+2) \cdot (n+3)\cdot(n-1)}$ $\large =~\color{#CEBB00}{n\cdot (n+1) \cdot (n+2) \cdot (n+3) \cdot (n+4}$ Hence, $A= \boxed{ \frac{\color{#CEBB00}{n\cdot(n+1)\cdot(n+2)\cdot(n+3)\cdot(n+4)}}{5}}$ Plugging in the value $n=10$ , we have $Ans~=~\frac{10 \cdot 11 \cdot 12 \cdot 13 \cdot 14}{5}=\LARGE \boxed{48048}$

This question deserves to be of level 3 but it has become level 2 because I think people are just adding the terms, I suggest Akhil Bansal to add more terms so that it gets back to level 3.

I will not post the actual answer because we already have from Kay Xspre

I will show some rigorous method:

1 2 3*4 = 24

2.3.4.5 = 120

3.4.5.6 = 360

4.5.6.7 = 840

Similary Carry on and add the result : 24 + 120 + 360 + 840 + ...... + 17160 = 48048