Chủ đề này có 2 trả lời

[sheep9]

a, $1+2+3+4+...+n= \frac{n(n+1)}{2}\forall n\in \mathbb{N}^{*}.$

b, $1^{2}+2^{2}+3^{2}+4^{2}+...+n^{2}=\frac{n(n+1)(2n+1)}{6}\forall n\in \mathbb{N}^{*}$

[Phuong Thu Quoc]

a/ $S=1+2+3+...+n$

    $S=n+\left ( n-1 \right )+\left ( n-2 \right )+...+1$

$\Rightarrow 2S=\left ( n+1 \right )+\left ( n+1 \right )+...+\left ( n+1 \right )=n\left ( n+1 \right )$

$\Rightarrow S=\frac{n\left ( n+1 \right )}{2}$

b/ $1^{3}=1$

    $2^{3}=\left ( 1+1 \right )^{3}=1^{3}+3.1^{2}+3.1+1$

    $3^{3}=\left ( 2+1 \right )^{3}=2^{3}+3.2^{2}+3.2+1$

    ......

    $\left ( n+1 \right )^{3}=n^{3}+3.n^{2}+3n+1$

Cộng vào $\Rightarrow \left ( 1^{3}+2^{3}+...+\left ( n+1 \right )^{3} \right )=\left ( 1^{3}+2^{3}+...+n^{3} \right )+3.\left ( 1^{2}+2^{2}+..+n^{2} \right )+3.\left ( 1+2+...+n \right )+\left ( 1+1+...+1 \right )$

$\Rightarrow \left ( n+1 \right )^{3}=3\sum_{k=1}^{n}k^{2}+3\sum_{k=1}^{n}k+\left ( n+1 \right )$

$\Rightarrow \sum_{k=1}^{n}k^{2}=\frac{n\left ( n+1 \right )\left ( 2n+1 \right )}{6}$

[sheep9]

a/ $S=1+2+3+...+n$

    $S=n+\left ( n-1 \right )+\left ( n-2 \right )+...+1$

$\Rightarrow 2S=\left ( n+1 \right )+\left ( n+1 \right )+...+\left ( n+1 \right )=n\left ( n+1 \right )$

$\Rightarrow S=\frac{n\left ( n+1 \right )}{2}$

b/ $1^{3}=1$

    $2^{3}=\left ( 1+1 \right )^{3}=1^{3}+3.1^{2}+3.1+1$

    $3^{3}=\left ( 2+1 \right )^{3}=2^{3}+3.2^{2}+3.2+1$

    ......

    $\left ( n+1 \right )^{3}=n^{3}+3.n^{2}+3n+1$

Cộng vào $\Rightarrow \left ( 1^{3}+2^{3}+...+\left ( n+1 \right )^{3} \right )=\left ( 1^{3}+2^{3}+...+n^{3} \right )+3.\left ( 1^{2}+2^{2}+..+n^{2} \right )+3.\left ( 1+2+...+n \right )+\left ( 1+1+...+1 \right )$

$\Rightarrow \left ( n+1 \right )^{3}=3\sum_{k=1}^{n}k^{2}+3\sum_{k=1}^{n}k+\left ( n+1 \right )$

$\Rightarrow \sum_{k=1}^{n}k^{2}=\frac{n\left ( n+1 \right )\left ( 2n+1 \right )}{6}$

em đang học lớp 10 nên không hiểu công thức cuối ạ