Chứng minh \(\frac{1}{2^2}\)+ \(\frac{1}{3^2}\)+ \(\frac{1}{4^2}\)+ .... + \(\frac{1}{2009^2}\)< 1
Chứng minh rằng \(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+.............+\frac{1}{2009\sqrt{2008}}< 2\)
Ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}\left(\frac{1}{\sqrt{n^2}}-\frac{1}{\sqrt{\left(n+1\right)^2}}\right)\)
\(=\sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(=\left(1+\frac{\sqrt{n}}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(< \left(1+1\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)=2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Áp dụng vào bài toán ta được
\(\frac{1}{2}+\frac{1}{3\sqrt{2}}+...+\frac{1}{2009\sqrt{2008}}\)
\(=2\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2008}}-\frac{1}{\sqrt{2009}}\right)\)
\(=2\left(1-\frac{1}{\sqrt{2009}}\right)< 2\)
Chứng minh rằng : \(1-\frac{1}{2^2}-\frac{1}{3^2}-...-\frac{1}{2009^2}>\frac{1}{2009}\)
Bài 1 : Cho N =\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2009^2}+\frac{1}{2010^2}\)
Hãy chứng minh rằng N<1
Xét N :
N = \(\frac{1}{2.2}\)+\(\frac{1}{3.3}\)+\(\frac{1}{4.4}\)+...+\(\frac{1}{2009.2009}\)+\(\frac{1}{2010.2010}\)
Ta có :
\(\frac{1}{2.2}\)< \(\frac{1}{1.2}\)
\(\frac{1}{3.3}\)< \(\frac{1}{2.3}\)
...
\(\frac{1}{2009.2009}\)<\(\frac{1}{2008.2009}\)
\(\frac{1}{2010.2010}\)<\(\frac{1}{2019.2010}\)
Cộng vế theo vế của các bất đẳng thức trên , ta có :
\(\frac{1}{2.2}\)+\(\frac{1}{3.3}\)+\(\frac{1}{4.4}\)+...+\(\frac{1}{2009.2009}\)+\(\frac{1}{2010.2010}\) < \(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+...+\(\frac{1}{2008.2009}\)+\(\frac{1}{2019.2010}\)
=> N < 1 - \(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+...+\(\frac{1}{2009}\)-\(\frac{1}{2010}\)
=> N < 1 - \(\frac{1}{2010}\)<1
=> N < 1
Chứng minh \(\frac{1}{2^2}\)+ \(\frac{1}{3^2}\)+ \(\frac{1}{4^2}\)+ .... + \(\frac{1}{2009^2}\)< 1
\(Cm:\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2009^2}< 1\)
Có : \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2009^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2008\cdot2009}\)
\(\Leftrightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2009^2}< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2008}-\frac{1}{2009}\)
\(\Leftrightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2009^2}< \frac{1}{1}-\frac{1}{2009}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2009^2}< \frac{2008}{2009}\left(1\right)\)
Mà \(\frac{2008}{2009}< 1\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2009^2}< \frac{2008}{2009}< 1\)
\(\Leftrightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2009^2}< 1\left(đpcm\right)\)
\(\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{2009^2}< \frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{2008.2009}\\ =\frac{1}{1}-\frac{1}{2009}< 1\left(\text{đ}pcm\right)\)
chứng minh rằng:
a.\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\).....+\(\frac{1}{2010^2}\)<1
b.\(\frac{1}{4}\)+\(\frac{1}{16}\)+\(\frac{1}{36}\)+\(\frac{1}{64}\)+\(\frac{1}{100}\)+\(\frac{1}{196}\)<\(\frac{1}{2}\)
c.\(\frac{2009^{2008}-1}{2009^{2009}-1}< \frac{2009^{2007}+1}{2009^{2008}+1}\)
a) Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}\) ; \(\frac{1}{3^2}< \frac{1}{2.3}\) ; \(\frac{1}{4^2}< \frac{1}{3.4}\) ; ... ; \(\frac{1}{2010^2}< \frac{1}{2009.2010}\)
=> \(Vt< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(=1-\frac{1}{2010}< 1\)
Chứng minh: \(1+\frac{1}{\sqrt[3]{2}}+\frac{1}{\sqrt[3]{3}}+\frac{1}{\sqrt[3]{4}}+...+\frac{1}{\sqrt[3]{2009}}>237\)
không dùng máy tính chứng minh \(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{2010\sqrt{2009}}< \frac{88}{45}\)
a) Chứng Minh Rằng : E = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2009^2}< 1\)
b) Tìm Các Số Nguyên n để : \(\frac{2n-1}{n+8}-\frac{n-14}{n+8}\)Là Số Nguyên
Chứng minh:
\(1+\frac{1}{\sqrt[3]{2}}+\frac{1}{\sqrt[3]{3}}+\frac{1}{\sqrt[3]{4}}+...+\frac{1}{\sqrt[3]{2009}}>237\)