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22 tháng 2 2016
Ta có:
\(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\)
\(=\frac{1}{4}+\left(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\right)\)
Đặt \(B=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\)
\(B=\left(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}\right)+\left(\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\right)\)
Giả sử tất cả các số hạng của B đều bằng \(\frac{1}{6^2}\)
\(\Rightarrow B=6.\frac{1}{6^2}=\frac{6}{36}=\frac{1}{6}<\frac{1}{4}\)
Do đó \(B<\frac{1}{4}\)
\(\Rightarrow A=\frac{1}{4}+B<\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\)
Vậy \(A<\frac{1}{2}\)
19 tháng 4 2016
Ta có : \(B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}\)
Mà \(\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3};...;\frac{1}{8^2}<\frac{1}{7.8}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}<\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}=1-\frac{1}{8}<1\)
Vậy B < 1
26 tháng 4 2016
Bạn xem lời giải của mình nhé:
Giải:
\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{6.7}+\frac{1}{7.8}\\\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8} \\ =\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(=1-\frac{1}{8}< 1\\ \Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}< 1\)
Chúc bạn học tốt!
HD
26 tháng 4 2016
Ta có: \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}< \frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{7.8}\)
= \(1-\frac{1}{8}< 1\)
Vậy B < 1
30 tháng 3 2016
B=3/2 xin loi nha
vì cách trình bày trên này khó quá, đọc chắc bạn ko hiểu đâu
22 tháng 3 2016
Ta có:
\(A=1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{10}}\)
\(=1-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
Đặt \(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
\(\Rightarrow2B=2\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
\(=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
\(\Rightarrow2B-B=B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
\(=1-\frac{1}{2^{10}}\)
\(\Rightarrow A=1-\left(1-\frac{1}{2^{10}}\right)=1-1+\frac{1}{2^{10}}=\frac{1}{2^{10}}\)
Vậy \(A=\frac{1}{2^{10}}\)
SY
19 tháng 3 2016
a)Đặt A= \(\frac{1}{2}\) - \(\frac{1}{4}\) + \(\frac{1}{8}\) - \(\frac{1}{16}\) + \(\frac{1}{32}\) - \(\frac{1}{64}\) => A=\(\frac{1}{2^1}\) - \(\frac{1}{2^2}\) + \(\frac{1}{2^3}\) - \(\frac{1}{2^4}\) + \(\frac{1}{2^5}\) - \(\frac{1}{2^6}\)
=> 2A= 1-\(\frac{1}{2^1}\) + \(\frac{1}{2^2}\) - \(\frac{1}{2^3}\) + \(\frac{1}{2^4}\) - \(\frac{1}{2^5}\)
=> 3A= 1- \(\frac{1}{2^6}\) <1 => A<\(\frac{1}{3}\) => đpcm.
b) Đặt B=\(\frac{1}{3}\) - \(\frac{2}{3^2}\) + \(\frac{3}{3^3}\) - \(\frac{4}{3^4}\) +..+ \(\frac{99}{3^{99}}\) - \(\frac{100}{3^{100}}\)
=> 3B=1-\(\frac{2}{3}\) + \(\frac{3}{3^2}\) - \(\frac{4}{3^3}\) +...+\(\frac{99}{3^{98}}\) - \(\frac{100}{3^{99}}\)
=> 4B= 1-\(\frac{1}{3}\) + \(\frac{1}{3^2}\) - \(\frac{1}{3^3}\) +...+\(\frac{1}{3^{99}}\) - \(\frac{100}{3^{99}}\) < 1-\(\frac{1}{3}\) + \(\frac{1}{3^2}\) - \(\frac{1}{3^3}\) +...+\(\frac{1}{3^{99}}\) (1)
Đặt B= 1-\(\frac{1}{3}\) + \(\frac{1}{3^2}\) - \(\frac{1}{3^3}\) +...+\(\frac{1}{3^{99}}\)
=> 3B= 3-1+\(\frac{1}{3}\) - \(\frac{1}{3^2}\) + \(\frac{1}{3^3}\) - \(\frac{1}{3^4}\) +...+ \(\frac{1}{3^{98}}\)
=> 4B= 3-\(\frac{1}{3^{99}}\) <3 => B<\(\frac{3}{4}\) (2)
=> 4A<B<\(\frac{3}{4}\) => A<\(\frac{3}{16}\) => đpcm.
17 tháng 4 2016
Ta thấy: \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+......+\frac{1}{50^2}\)<\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{49.50}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{50^2}\)<\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{49}-\frac{1}{50}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{50^2}\)<\(1-\frac{1}{50}\)
Suy ra:
A=\(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{50^2}\)<\(\frac{1}{1^2}+\left(1-\frac{1}{50}\right)\)
A<1+1-\(\frac{1}{50}\)
A<2-\(\frac{1}{50}\)<2
Vậy A<2(đpcm)
10 tháng 7 2019
\(Q=\frac{2\cdot2010}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+2012}}\)
\(Q=\frac{2\cdot2010}{1+\frac{1}{\frac{(1+2)\cdot2}{2}}+\frac{1}{\frac{(1+3)\cdot3}{2}}+\frac{1}{\frac{(1+4)\cdot4}{2}}+...+\frac{1}{\frac{(1+2012)\cdot2012}{2}}}\)
\(Q=\frac{2\cdot2010}{1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{2025078}}\)
\(Q=\frac{2\cdot2010}{1+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}...+\frac{2}{4050156}}\)
\(Q=\frac{2\cdot2010}{1+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}+...+\frac{2}{2012\cdot2013}}\)
\(Q=\frac{2\cdot2010}{1+2\left[\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2012}-\frac{1}{2013}\right]}\)
\(Q=\frac{2\cdot2010}{1+2\left[\frac{1}{2}-\frac{1}{2013}\right]}=\frac{2\cdot2010}{1+\frac{2011}{2013}}=\frac{2\cdot2010}{\frac{4024}{2013}}=\frac{4020}{\frac{4024}{2013}}=4020\cdot\frac{2013}{4024}=...\)
Nguyễn Linh Chi ơi , hình như cô nhầm thì phải :v \(2-\frac{2}{2013}=\frac{2\cdot2013-2}{2013}=\frac{4026-2}{2013}=\frac{4024}{2013}\)
sao mà bằng \(\frac{4020}{2013}\)được cô
10 tháng 7 2019
Ta có:
\(P=1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2012}\)
\(P=1+\frac{1}{\frac{\left(1+2\right).2}{2}}+\frac{1}{\frac{\left(1+3\right).3}{2}}+...+\frac{1}{\frac{\left(1+2012\right).2012}{2}}\)
\(P=1+\frac{2}{3.2}+\frac{2}{4.3}+...+\frac{2}{2013.2012}\)
\(P=1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2012}-\frac{1}{2013}\right)\)\
\(P=1+2\left(\frac{1}{2}-\frac{1}{2013}\right)\)
\(P=1+1-\frac{2}{2013}=2-\frac{2}{2013}=\frac{4020}{2013}\)
\(Q=\frac{2.2010}{P}=\frac{4020}{\frac{4020}{2013}}=2013\)....
Ta có :
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}=\frac{9}{10}<\frac{10}{10}=1\)
Có : \(\frac{1}{2^2}<1\)
\(\frac{1}{3^2}<1\)
\(\frac{1}{4^2}<1\)
...
\(\frac{1}{10^2}<1\)
Cộng tất cả các vế trên ta có: \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}<1\) (ĐPCM)