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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
\(\sqrt{1+2+3+...+\left(n-1\right)+n+\left(n-1\right)+...+3+2+1}\\ =\sqrt{2\left[1+2+3+...+\left(n-1\right)+n\right]-n}\\ =\sqrt{2.\left(n+1\right).n:2-n}\\ =\sqrt{n\left(n+1\right)-n}\\ =\sqrt{n^2+n-n}\\ =\sqrt{n^2}\\ =n\)
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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a: \(\left(n^2+3n-1\right)\left(n+2\right)-n^3+2\)
\(=n^3+2n^2+3n^2+6n-n-2+n^3+2\)
\(=5n^2+5n=5\left(n^2+n\right)⋮5\)
b: \(\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-1\right)\)
\(=6n^2+30n+n+5-6n^2+3n-10n+5\)
\(=24n+10⋮2\)
d: \(=\left(n+1\right)\left(n^2+2n\right)\)
\(=n\left(n+1\right)\left(n+2\right)⋮6\)
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
Thiếu điều kiên n E N
\(\sqrt{1+2+3...+\left(n-1\right)+n+\left(n-1\right)+...+3+2+1}\)
\(=\sqrt{2\left[1+2+3+..+\left(n-1\right)+n\right]}=\sqrt{2\frac{n\left(n-1\right)}{2}+n}\)
\(=\sqrt{n\left(n-1\right)+n}=\sqrt{n^2-n+n}=\sqrt{n^2}=n\left(đpcm\right)\)