Ta có :

\(\sqrt{1+2+...+n-1+n+n-1+...+2+1}\)

=\(\sqrt{2\left(1+2+...+n-1\right)+n}\)

=\(\sqrt{\dfrac{2\left(n-1\right)n}{2}+n}=\sqrt{n^2}=n\)

Chúc Bạn Học Tốt ,Cô @Bùi Thị Vân kiểm tra giùm em với ạ

Thiếu  điều  kiên n E N

\(\sqrt{1+2+3...+\left(n-1\right)+n+\left(n-1\right)+...+3+2+1}\)

\(=\sqrt{2\left[1+2+3+..+\left(n-1\right)+n\right]}=\sqrt{2\frac{n\left(n-1\right)}{2}+n}\)

\(=\sqrt{n\left(n-1\right)+n}=\sqrt{n^2-n+n}=\sqrt{n^2}=n\left(đpcm\right)\)

Ta có:

\(\sqrt{1+2+...+n-1+n+n-1+...+2+1}\)

\(=\sqrt{2\left(1+2+...+n-1\right)+n}\)

\(=\sqrt{\frac{2\left(n-1\right)n}{2}+n}=\sqrt{n^2}=n\)

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Đặt \(A_k=1+2+3+4+.....+k=\frac{k\left(k+1\right)}{2}\Rightarrow A_k^2=\frac{k^2\left(k+1\right)^2}{4}\)

\(A_{k-1}=1+2+3+4+.....+\left(k-1\right)=\frac{k\left(k-1\right)}{2}\Rightarrow A_{k-1}^2=\frac{k^2\left(k-1\right)^2}{4}\)

\(\Rightarrow A_k^2-A_{k-1}^2=\frac{k^2\left(k+1\right)^2-k^2\left(k-1\right)^2}{4}=\frac{k^2\left(k^2+2k+1-k^2+2k-1\right)}{4}=\frac{4k^3}{4}=k^3\)

Khi đó:

\(1^3=A_1^2\)

\(2^3=A_2^2-A_1^2\)

\(3^3=A_3^2-A_2^2\)

\(.........................................................................................\)

\(n^3=A_n^2-A_{n-1}^2\)

\(\Rightarrow1^3+2^3+3^3+.....+n^3=A_n^2=\left(1+2+3+......+n\right)^2=\left[\frac{n\left(n+1\right)}{2}\right]^2\)

Đề ghi sót . Vế cuối là móc vuông đó bình phương chư

Vì \(n\in Z^+\)nên\(n\left(n+1\right)\left(n+2\right)>n^3\Rightarrow\sqrt[3]{n\left(n+1\right)\left(n+2\right)}>n\)

\(\Rightarrow\sqrt[3]{n\left(n+1\right)\left(n+2\right)+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}+...+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}}>n\)(1)

Lại có:\(n^2+2n+1>n^2+2n\Rightarrow\left(n+1\right)^2>n\left(n+2\right)\Rightarrow\left(n+1\right)^3>n\left(n+1\right)\left(n+2\right)\)

\(\Rightarrow n+1>\sqrt[3]{n\left(n+1\right)\left(n+2\right)}\\ \Rightarrow\sqrt[3]{n^3+3n^2+3n+1}>\sqrt[3]{n^3+3n^2+2n}\)

\(\Rightarrow\sqrt[3]{n^3+3n^2+2n+n+1}>\sqrt[3]{n^3+3n^2+2n+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}}\)

\(\Rightarrow\sqrt[3]{\left(n+1\right)^3}>\sqrt[3]{n\left(n+1\right)\left(n+2\right)+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}}\)

Tương tự \(\Rightarrow n+1>\sqrt[3]{n\left(n+1\right)\left(n+2\right)+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}+...+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}}\)(2)

Từ (1) và (2) suy ra:

\(n< \sqrt[3]{n\left(n+1\right)\left(n+2\right)+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}+...+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}}< n+1\)

\(n\in Z^+\)nên n² < n² + 2n < n² + 2n + 1 <=> n² < n(n + 2) < (n + 1)² => n³ < n(n + 1)(n + 2) < (n + 1)³ 

=>\(n< \sqrt[3]{n\left(n+1\right)\left(n+2\right)}< n+1\)

=>\(n< \sqrt[3]{n\left(n+1\right)\left(n+2\right)}< \sqrt[3]{n\left(n+1\right)\left(n+2\right)+n}\)\(< \sqrt[3]{n\left(n+1\right)\left(n+2\right)+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}}< \sqrt[3]{n\left(n+1\right)\left(n+2\right)+n+1}\)\(=\sqrt[3]{\left(n+1\right)\left(n^2+2n+1\right)}=\sqrt[3]{\left(n+1\right)\left(n+1\right)^2}=n+1\)

=>\(n< \sqrt[3]{n\left(n+1\right)\left(n+2\right)+n}\)

\(< \sqrt[3]{n\left(n+1\right)\left(n+2\right)+\sqrt[3]{n\left(n+1\right)\left(n+2\right)+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}}}< n+1\)

Tiếp tục như vậy,ta có đpcm.

Sorry ! n² < n(n + 2) nên n³ < n(n + 1)(n + 2) (vì n < n + 1)

vd:n=-0,8 thì sai

Chứng minh 

\(\sqrt[3]{\left(n+1\right)^2}-\sqrt[3]{n^2}< \frac{2}{3\sqrt[3]{n}}\)

\(\Leftrightarrow3\sqrt[3]{n\left(n+1\right)^2}< 2+3n\)

Lập phương 2 vế rồi rút gọn được

\(\Leftrightarrow9n+8>0\)

Đúng với mọi n dương. Ta có ĐPCM.

Cái còn lại tương tự