a, $1+2+3+4+...+n= \frac{n(n+1)}{2}\forall n\in \mathbb{N}^{*}.$
b, $1^{2}+2^{2}+3^{2}+4^{2}+...+n^{2}=\frac{n(n+1)(2n+1)}{6}\forall n\in \mathbb{N}^{*}$
a, $1+2+3+4+...+n= \frac{n(n+1)}{2}\forall n\in \mathbb{N}^{*}.$
b, $1^{2}+2^{2}+3^{2}+4^{2}+...+n^{2}=\frac{n(n+1)(2n+1)}{6}\forall n\in \mathbb{N}^{*}$
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 $\Rightarrow \left ( 1^{3}+2^{3}+...+\left ( n+1 \right )^{3} \right )=\left ( 1^{3}+2^{3}+...+n^{3} \right )+3.\left ( 1^{2}+2^{2}+..+n^{2} \right )+3.\left ( 1+2+...+n \right )+\left ( 1+1+...+1 \right )$
$\Rightarrow \left ( n+1 \right )^{3}=3\sum_{k=1}^{n}k^{2}+3\sum_{k=1}^{n}k+\left ( n+1 \right )$
$\Rightarrow \sum_{k=1}^{n}k^{2}=\frac{n\left ( n+1 \right )\left ( 2n+1 \right )}{6}$
Thà một phút huy hoàng rồi chợt tối
Còn hơn buồn le lói suốt trăm năm.
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
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