Tính `1^2`+`2^2`+...+`n^2`

Đáp án:

$1^2+2^2\ +\,.\!.\!.+\ n^2=\dfrac{n(n+1)(2n-1)}6$.

Giải thích các bước giải:

Cho biểu thức là A.

$A=1^2+2^2\ +\,.\!.\!.+\ n^2\\\Rightarrow A=1.1+2.2\ +\,.\!.\!.+\ n.\!n\\\Rightarrow A=1.(2-1)+2.(3-1)\ +\,.\!.\!.+\ n[(n+1)-1]\\\Rightarrow A=1.2-1+2.3-2\ +\,.\!.\!.+\ n(n+1)-n\\\Rightarrow A=\bigg(\underbrace{1.2+2.3\ +\,.\!.\!.+\ n(n+1)}_{\large A'}\bigg)-\bigg(\underbrace{1+2\ +\,.\!.\!.+\ n}_{\large A''}\bigg)\\A'=1.2+2.3\ +\,.\!.\!.+\ n(n+1)\\\Rightarrow 3A'=1.2.3+2.3.3\ +\,.\!.\!.+\ n(n+1).3\\\Rightarrow 3A'=1.2.3+2.3.(4-1)\ +\,.\!.\!.+\ n(n+1)[(n+2)-(n-1)]\\\Rightarrow 3A'=1.2.3+2.3.4-1.2.3\ +\,.\!.\!.+\ n(n+1)(n+2)-(n-1)n(n+1)\\\Rightarrow 3A'=(1.2.3+2.3.4\ +\,.\!.\!.+\ n(n+1)(n+2))-(1.2.3\ +\,.\!.\!.+\ (n-1)n(n+1))\\\Rightarrow 3A'=n(n+1)(n+2)\\\Rightarrow A'=\dfrac{n(n+1)(n+2)}3\\\Rightarrow A''=1+2\ +\,.\!.\!.+\ n\quad|\quad(n-1):1+1=n(s/h)\\\Rightarrow A''=(1+n)+[2+(n-1)]\ +\,.\!.\!.\\\Rightarrow A''=(n+1)+(n+1)\ +\,.\!.\!.\\\Rightarrow A''=(n+1)\!\cdot\!\dfrac n2\\\Rightarrow A''=\dfrac{(n+1).\!n}2\\\Rightarrow A=A'-A''=\dfrac{n(n+1)(n+2)}3-\dfrac{n(n+1)}2\\\Rightarrow A=\dfrac{2n(n+1)(n+2)}6-\dfrac{3n(n+1)}6\\\Rightarrow A=\dfrac{2n(n+1)(n+2)-3n(n+1)}6\\\Rightarrow A=\dfrac{n(n+1)[2(n+2)-3]}6\\\Rightarrow A=\dfrac{n(n+1)(2n+4-3)}6\\\Rightarrow A=\dfrac{n(n+1)(2n-1)}6$

Vậy $1^2+2^2\ +\,.\!.\!.+\ n^2=\dfrac{n(n+1)(2n-1)}6$.