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a)= \(\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+...+\frac{\sqrt{100}-\sqrt{99}}{100-99}\)
=\(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}\)
= \(-1+\sqrt{100}\)
= -1 +10
=9
b)Ta có\(\left(\sqrt{n+1}-\sqrt{n}\right)\cdot\left(\sqrt{n+1}+\sqrt{n}\right)\)=n+1-n=1 (1)
Lại có:\(\frac{1}{\sqrt{n+1}+1}\cdot\left(\sqrt{n+1}+1\right)=1\)(2)
Từ (1) và (2)=>\(\left(\sqrt{n+1}-1\right)=\frac{1}{\sqrt{n+1}+1}\)
Xét vế trái : \(\left(\sqrt{n+1}-\sqrt{n}\right)^2=2n+1-2\sqrt{n}.\sqrt{n+1}\)
Xét vế phải : \(\sqrt{\left(2n+1\right)^2}-\sqrt{\left(2n+1\right)^2-1}=\left|2n+1\right|-\sqrt{\left(2n+1-1\right)\left(2n+1+1\right)}\)
\(=2n+1-\sqrt{2n.2\left(n+1\right)}=2n+1-2\sqrt{n}.\sqrt{n+1}\)
=> VT = VP => đpcm
\(\sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}}\)
VP = \(\frac{1}{\sqrt{n+1}+\sqrt{n}}\)
= \(\frac{\sqrt{n+1}-\sqrt{n}}{\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}\)
= \(\frac{\sqrt{n+1}-\sqrt{n}}{\left(\sqrt{n+1}\right)^2-\left(\sqrt{n}\right)^2}\)
= \(\frac{\sqrt{n+1}-\sqrt{n}}{n+1-n}\)
= \(\sqrt{n+1}-\sqrt{n}\)
= VT
Vậy đẳng thức được chứng minh
\(\frac{1}{\sqrt{n+1}+\sqrt{n}}\) = \(\frac{\sqrt{1+n}-\sqrt{n}}{1+n-n}=\:\sqrt{n+1}-\sqrt{n}\)
\(\sqrt{n+1}-\sqrt{n}=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\sqrt{n+1}+\sqrt{n}}=\frac{\left(\sqrt{n+1}\right)^2-\left(\sqrt{n}\right)^2}{\sqrt{n+1}+\sqrt{n}}=\frac{n+1-n}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}\)
Ta có : \(\frac{1}{2\sqrt{n+1}}=\frac{1}{\sqrt{n+1}+\sqrt{n+1}}< \frac{1}{\sqrt{n}+\sqrt{n+1}}=\frac{\sqrt{n+1}-\sqrt{n}}{\left(n+1\right)-n}=\sqrt{n+1}-\sqrt{n}\)
\(\Rightarrow\frac{1}{2\sqrt{n+1}}< \sqrt{n+1}-\sqrt{n}\)
Áp dụng : \(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2500}}=2\left(\frac{1}{2\sqrt{1}}+\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}+...+\frac{1}{2\sqrt{2500}}\right)< 2\left(1+\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{2500}-\sqrt{2499}\right)=2\sqrt{2500}=2.50=100\)
Vậy ta có điều phải chứng minh.
- Ta xét : \(\frac{1}{\sqrt{n}}=\frac{2}{\sqrt{n}+\sqrt{n}}>\frac{2}{\sqrt{n}+\sqrt{n+1}}=\frac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{\left(n+1\right)-n}=2\left(\sqrt{n+1}-\sqrt{n}\right)< 2\sqrt{n+1}-2\)
- Ta xét : \(\frac{1}{\sqrt{n}}=\frac{2}{\sqrt{n}+\sqrt{n}}< \frac{2}{\sqrt{n}+\sqrt{n-1}}=\frac{2\left(\sqrt{n}-\sqrt{n-1}\right)}{n-\left(n-1\right)}=2\left(\sqrt{n}-\sqrt{n-1}\right)< 2\sqrt{n}\) ;
\(\sqrt{n+1}-\sqrt{n}=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\sqrt{n+1}+\sqrt{n}}=\frac{\left(n+1\right)-n}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}\) (đpcm)