Question

[Huỳnh Thanh Xuân]

chứng minh rằng

a) \(\sqrt{4+\sqrt{4+....+\sqrt{4}+\sqrt{ }4}}< 3\)

b)\(\sqrt{1^3+2^3+...+n^3}=1+2+...+n\)

Answer 1

[Nhật Minh]

\(A=\sqrt{4+\sqrt{4+\sqrt{4}+...}}\\ \)>0

a)

\(A=\sqrt{4+A}\Leftrightarrow A^2=4+A\Leftrightarrow A^2-A-4=0\)

\(\Delta=1+16=17\)

\(A_1=\dfrac{1+\sqrt{17}}{2}< \dfrac{1+5}{2}=3\)

\(A_2=\dfrac{1-\sqrt{17}}{2}\)<0 loại

Vậy A < 3

b) Chứng minh quy nạp

(1³+2³+.....+n³)=(1+2+3+...+n)²=> KL

Answer 2

[Neet]

b).đặt \(A=\sqrt{1^3+2^3+3^3+...+n^3}\)

ta có hằng đẳng thức: \(x^3-x=\left(x-1\right)x\left(x+1\right)\)

\(1^3+2^3+3^3+...+n^3=1^3-1+2^3-2+3^3-3+...+n^3-n+\left(1+2+3+...+n\right)\)\(=0+1.2.3+2.3.4+...+\left(n-1\right)n\left(n+1\right)+\dfrac{n\left(n+1\right)}{2}\)(*)

Xét \(B=1.2.3+2.3.4+...+\left(n-1\right)n\left(n+1\right)\)

\(4B=1.2.3.4+2.3.4.4+...+\left(n-1\right)n\left(n+1\right).4=1.2.3.4+2.3.4.5-1.2.3.4+...+\left(n-1\right)n\left(n+1\right)\left(n+2\right)-\left(n-2\right)\left(n-1\right)n\left(n+1\right)\)

\(=\left(n-1\right)n\left(n+1\right)\left(n+2\right)\)

\(\Rightarrow B=\dfrac{\left(n-1\right)n\left(n+1\right)\left(n+2\right)}{4}\)

từ (*): \(1^3+2^3+...+n^3=\dfrac{\left(n-1\right)n\left(n+1\right)\left(n+2\right)}{4}+\dfrac{n\left(n+1\right)}{2}\)

\(=\dfrac{n\left(n+1\right)}{2}\left[\dfrac{\left(n-1\right)\left(n+2\right)}{2}+1\right]=\dfrac{n\left(n+1\right)}{2}.\dfrac{n^2+n-2+2}{2}=\left[\dfrac{n\left(n+1\right)}{2}\right]^2\)

do đó \(A=\sqrt{\left[\dfrac{n\left(n+1\right)}{2}\right]^2}=\dfrac{n\left(n+1\right)}{2}=1+2+...+n\)(đpcm)