Chứng minh \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\) rồi áp dụng với n = 1,2,....,2014

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Ta có:

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

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

Thế vô bài toán được

\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{2016\sqrt{2015}+2015\sqrt{2016}}\)

\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}\)

\(=1-\frac{1}{\sqrt{2016}}\)

a,a=b+1

suy ra a-b=1 suy ra(\(\sqrt{a}+\sqrt{b}\))(\(\sqrt{a}-\sqrt{b}\))=1

suy ra \(\sqrt{a}-\sqrt{b}\)=\(\frac{1}{\sqrt{a}+\sqrt{b}}\)(1)

vì a=b+1 suy ra a>b suy ra \(\sqrt{a}>\sqrt{b}\)suy ra \(\sqrt{a}+\sqrt{b}>2\sqrt{b}\)

suy ra \(\frac{1}{\sqrt{a}+\sqrt{b}}< \frac{1}{2\sqrt{b}}\)(2)

từ (1) ,(2) suy ra\(\sqrt{a}-\sqrt{b}< \frac{1}{2\sqrt{b}}\)suy ra \(2\left(\sqrt{a}-\sqrt{b}\right)< \frac{1}{\sqrt{b}}\)(*)

ta lại có b+1=c+2 suy ra b-c =1 suy ra\(\left(\sqrt{b}-\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)=1\)

suy ra \(\sqrt{b}-\sqrt{c}=\frac{1}{\sqrt{b}+\sqrt{c}}\)(3)

vì b>c suy ra \(\sqrt{b}>\sqrt{c}\) suy ra \(\sqrt{b}+\sqrt{c}>2\sqrt{c}\)

suy ra \(\frac{1}{\sqrt{b}+\sqrt{c}}< \frac{1}{2\sqrt{c}}\)(4)

Từ (3),(4) suy ra \(\sqrt{b}-\sqrt{c}< \frac{1}{2\sqrt{c}}\) suy ra\(2\left(\sqrt{b}+\sqrt{c}\right)< \frac{1}{\sqrt{c}}\)(**)

từ (*),(**) suy ra đccm

Mỗi biểu thức trong dấu căn có dạng:

\(1+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}\)   ( Với \(k\ge2\))

Ta có:

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

\(=\frac{k^4+2k^2\left(k+1\right)+\left(k+1\right)^2}{k^2\left(k+1\right)^2}=\frac{\left(k^2+k+1\right)^2}{\left(k\left(k+1\right)\right)^2}\)

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

\(\Rightarrow S=1+1-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}-\frac{1}{4}+...+1+\frac{1}{2013}-\frac{1}{2014}=2014-\frac{1}{2014}\)

Mỗi biểu thức trong dấu căn có dạng:

1+1k2 +1(k+1)2    ( Với k≥2)

Ta có:

1+1k2 +1(k+1)2 =k2(k+1)2+(k+1)2+k2k2(k+1)2 =k4+2k3+k2+k2+2k+1+k2k2(k+1)2 

=k4+2k2(k+1)+(k+1)2k2(k+1)2 =(k2+k+1)2(k(k+1))2 

⇒√1+1k2 +1(k+1)2 =k2+k+1k2+k =1+1k(k+1) =1+1k −1k+1 

⇒S=1+1−12 +1+12 −13 +1+13 −14 +...+1+12013 −12014 =2014−12014 

Mỗi biểu thức trong dấu căn có dạng:

1+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}1+k21​+(k+1)21​   ( Với k\ge2k≥2)

Ta có:

1+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}=\frac{k^2\left(k+1\right)^2+\left(k+1\right)^2+k^2}{k^2\left(k+1\right)^2}=\frac{k^4+2k^3+k^2+k^2+2k+1+k^2}{k^2\left(k+1\right)^2}1+k21​+(k+1)21​=k2(k+1)2k2(k+1)2+(k+1)2+k2​=k2(k+1)2k4+2k3+k2+k2+2k+1+k2​

=\frac{k^4+2k^2\left(k+1\right)+\left(k+1\right)^2}{k^2\left(k+1\right)^2}=\frac{\left(k^2+k+1\right)^2}{\left(k\left(k+1\right)\right)^2}=k2(k+1)2k4+2k2(k+1)+(k+1)2​=(k(k+1))2(k2+k+1)2​

\Rightarrow\sqrt{1+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}}=\frac{k^2+k+1}{k^2+k}=1+\frac{1}{k\left(k+1\right)}=1+\frac{1}{k}-\frac{1}{k+1}⇒1+k21​+(k+1)21​​=k2+kk2+k+1​=1+k(k+1)1​=1+k1​−k+11​

\Rightarrow S=1+1-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}-\frac{1}{4}+...+1+\frac{1}{2013}-\frac{1}{2014}=2014-\frac{1}{2014}⇒S=1+1−21​+1+21​−31​+1+31​−41​+...+1+20131​−20141​=2014−

\(\frac{1}{\sqrt{2}-\sqrt{3}}-\frac{1}{\sqrt{3}-\sqrt{4}}+...-\frac{1}{\sqrt{2013}-\sqrt{2014}}+\frac{1}{\sqrt{2014}-\sqrt{2015}}\)

\(=\frac{\sqrt{2}+\sqrt{3}}{2-3}-\frac{\sqrt{3}+\sqrt{4}}{3-4}+...+\frac{\sqrt{2014}+\sqrt{2015}}{2014-2015}\)

\(=-\left(\sqrt{2}+\sqrt{3}\right)+\sqrt{3}+\sqrt{4}-\left(\sqrt{4}+\sqrt{5}\right)+...+\sqrt{2014}+\sqrt{2015}\)

=\(-\sqrt{2}+\sqrt{2015}\)