Cho tổng A = \(\frac{1}{2^2}\)+ \(\frac{1}{3^2}\)+...+ \(\frac{1}{2011^2}\). Chứng tỏ A < 1
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\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2011^2}\)
\(\text{Vì}\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};...;\frac{1}{2011^2}< \frac{1}{2010.2011}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2011^2}< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2010.2011}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2011^2}< \frac{1}{2^2}+\frac{1}{2}-\frac{1}{2011}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2011^2}< \frac{1}{4}+\frac{1}{2}-\frac{1}{2011}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2011^2}< \frac{3}{4}-\frac{1}{2011}< \frac{3}{4}\)
\(\Rightarrowđpcm\)
Ta có :
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2011^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2010.2011}\)\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2010}-\frac{1}{2011}=1-\frac{1}{2011}=\frac{2010}{2011}>\frac{2010}{2680}=\frac{3}{4}\)
Hình như có gì đó sai sai :')
Ta có: \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2010^2}<\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2009.2010}\)
\(<1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(<1-\frac{1}{2010}\)
\(<\frac{2009}{2010}<1\)
=>N<1
c)\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2012}}\)
\(2A=2\left(1+\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^{2012}}\right)\)
\(2A=2+1+\frac{1}{2^2}+\frac{1}{2^3}+.....+\frac{1}{2^{2011}}\)
\(2A-A=\left(2+1+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....\frac{1}{2^{2012}}\right)\)
\(A=2-\frac{1}{2^{2012}}\)
1/
A=1/1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
A=1/1-1/100
Vì 1/100>0
-->1/1-1/100<1
-->A<1
Bạn xem lời giải của mình nhé:
Giải:
A luôn > 0 (vì các số hạng trong tổng A đều lớn hơn 0)(1)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\\ 2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\\ 2A-A=1-\frac{1}{2^{100}}< 1\)
\(A< 1\)(2)
Từ (1) và (2) \(\Rightarrow0< A< 1\left(đpcm\right)\)
Chúc bạn học tốt!
\(n^2>\left(n-1\right)\left(n+1\right)\Rightarrow\frac{1}{n^2}< \frac{1}{\left(n-1\right)\left(n+1\right)}=\frac{1}{2}\left(\frac{1}{n-1}-\frac{1}{n+1}\right).\)
Do đó: \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2013^2}+\frac{1}{2014^2}< \frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{2012.2014}+\frac{1}{2013.2015}=\)
\(=\frac{1}{2}[1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2012}-\frac{1}{2014}+\frac{1}{2013}-\frac{1}{2015}]=\)
\(=\frac{1}{2}[1+\frac{1}{2}-\frac{1}{2014}-\frac{1}{2015}]=\frac{1}{2}[\frac{3}{2}-\frac{1}{2014}-\frac{1}{2015}]=\frac{3}{4}-\frac{1}{2}\left(\frac{1}{2014}+\frac{1}{2015}\right)< \frac{3}{4}.\)
Đặt \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2010.2011}\)
Ta có:
\(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2011^2}\)\(< \)\(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2010.2011}\left(1\right)\)
Mà \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2010.2011}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2010}-\frac{1}{2011}\)
\(=1-\frac{1}{2011}< 1\left(2\right)\)
Từ (1) và (2) \(\Rightarrow A< B< 1\Rightarrow A< 1\)
Đpcm