Đáp án:

$C=\dfrac{17.(n-1).n.(n+1).(n+2)}{12}$

Giải thích các bước giải:

$C=1^3+2^3+3^3\ +\,.\!.\!.+\ n^3 \\\Rightarrow C=1.1.1+2.2.2+3.3.3\ +\,.\!.\!.+\ n.n.n \\\Rightarrow C=1.1.(2-1)+2.2.(3-1)+3.3.(4-1)\ +\,.\!.\!.+\ n.n.(n+1-1) \\\Rightarrow C=1.1.2-1.1+2.2.3-2.2+3.3.4-3.3+...+n.n.(n+1)-n.n \\\Rightarrow C=\bigg(\underbrace{1.1.2+2.2.3+3.3.4+...+n.n.(n+1)}_{\large C'}\bigg)-\bigg(\underbrace{1.1+2.2+3.3+...+n.n}_{\large C''}\bigg) \\\Rightarrow C'=1.1.2+2.2.3+3.3.4+...+n.n.(n+1) \\\Rightarrow C'=(0+1).1.2+(1+1).2.3+(2+1).3.4+...+(n-1+1).n.(n+1) \\\Rightarrow C'=0.1.2+1.2+1.2.3+2.3+2.3.4+3.4+...+(n-1).n.(n+1)+n.(n+1) \\\Rightarrow C'=\bigg(\underbrace{0.1.2+1.2.3+2.3.4+...+(n-1).n.(n+1)}_{\large C'''}\bigg)+\bigg(\underbrace{1.2+2.3+3.4+...+n.(n+1)}_{\large C''''}\bigg) \\\Rightarrow C'''=0.1.2+1.2.3+2.3.4\ +\,.\!.\!.+\ (n-1).n.(n+1)\\\Rightarrow 4C'''=1.2.3.4+2.3.4.4\ +\,.\!.\!.+\ (n-1).n.(n+1).4\\\Rightarrow 4C'''=1.2.3.4+2.3.4.(5-1)\ +\,.\!.\!.+\ (n-1).n.(n+1).[(n+2)-(n-2)]\\\Rightarrow 4C'''=1.2.3.4+2.3.4.5-1.2.3.4\ +\,.\!.\!.+\ (n-1).n.(n+1).(n+2)-(n-2).(n-1).n.(n+1)\\\Rightarrow 4C'''=(1.2.3.4+2.3.4.5\ +\,.\!.\!.+\ (n-1).n.(n+1).(n+2))-(1.2.3.4+2.3.4.5\ +\,.\!.\!.+\ (n-2).(n-1).n.(n+1))\\\Rightarrow 4C'''=(n-1).n.(n+1).(n+2)\\\Rightarrow C'''=\dfrac{(n-1).n.(n+1).(n+2)}{4}\\\Rightarrow C''''=1.2+2.3+3.4\ +\,.\!.\!.+\ n.(n+1)\\\Rightarrow 3C''''=1.2.3+2.3.3+3.4.3\ +\,.\!.\!.+\ n.(n+1).3\\\Rightarrow 3C''''=1.2.3+2.3.(4-1)+3.4.(5-2)\ +\,.\!.\!.+\ n. (n+1).[(n+2)-(n-1)]\\\Rightarrow 3C''''=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4\ +\,.\!.\!.+\ n.(n+1).(n+2)-(n-1).n.(n+1)\\\Rightarrow 3C''''=(1.2.3+2.3.4+3.4.5\ +\,.\!.\!.+\ n.(n+1).(n+2))-(1.2.3+2.3.4\ +\,.\!.\!.+\ (n-1).n.(n+1)\\\Rightarrow 3C''''=n.(n+1).(n+2)\\\Rightarrow C''''=\dfrac{n.(n+1).(n+2)}{3}\\\Rightarrow C'=\dfrac{(n-1).n.(n+1).(n+2)}{4}+\dfrac{n.(n+1).(n+2)}{3}\\\Rightarrow C'=\dfrac{7.(n-1).n.(n+1).(n+2)}{12}\\\Rightarrow C''=1.1+2.2+3.3\ +\,.\!.\!.+\ n.n\\\Rightarrow C''=1+2.(1+1)+3.(2+1)\ +\,.\!.\!.+\ n.(n-1+1)\\\Rightarrow C''=1+1.2+2+2.3+3\ +\,.\!.\!.+\ (n-1).n+n\\\Rightarrow C''=\bigg(\underbrace{1.2+2.3\ +\,.\!.\!.+\ (n-1).n}_{\large C,}\bigg)+\bigg(1+2+3\ +\,.\!.\!.+ \ n\bigg)\\\Rightarrow C,=1.2+2.3\ +\,.\!.\!.+\ (n-1).n\\\Rightarrow 3C,=1.2.3+2.3.3\ +\,.\!.\!.+\ (n-1).n.3\\\Rightarrow 3C,=1.2.3+2.3.(4-1)\ +\,.\!.\!.+\ (n-1).n.[(n+1)-(n-2)]\\\Rightarrow 3C,=1.2.3+2.3.4-1.2.3\ +\,.\!.\!.+\ (n-1).n.(n+1)-(n-2).(n-1).n\\\Rightarrow 3C,=(1.2.3+2.3.4\ +\,.\!.\!.+\ (n-1).n.(n+1))-(1.2.3+2.3.4\ +\,.\!.\!.+\ (n-2).(n-1).n\\\Rightarrow 3C,=(n-1).n.(n+1)\\\Rightarrow C,=\dfrac{(n-1).n.(n+1)}{3}\\\Rightarrow C''=\dfrac{(n-1).n.(n+1)}{3}+\dfrac{n.(n+1)}{2}\\\Rightarrow C''=\dfrac{5.(n-1).n.(n+1)}{6}\\\Rightarrow C=\dfrac{7.(n-1).n.(n+1).(n+2)}{12}+\dfrac{5.(n-1).n.(n+1)}{6}\\\Rightarrow C=\dfrac{17.(n-1).n.(n+1).(n+2)}{12}$