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1 tháng 8 2017
S=\(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{2010.2011.2012}\)
=\(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2010.2011}-\frac{1}{2011.2012}\)
=\(\frac{1}{2}-\frac{1}{2011.2012}< \frac{1}{2}\)(Vì \(\frac{1}{2011.2012}>0\))
=> S <\(\frac{1}{2}\)
DP
2 tháng 8 2017
\(S=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+....+\frac{2}{2010.2011.2012}\)
\(S=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{2012-2010}{2010.2011.2012}\)
\(S=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2010.2011}-\frac{1}{2011.2012}\)
\(S=\frac{1}{1.2}-\frac{1}{2011.2012}=\frac{2023065}{4046132}\)
\(\text{Vì}\)\(\frac{2023065}{4046132}< \frac{1}{2}\Rightarrow S< P\)
22 tháng 7 2020
Bài 15 :
a) Đặt \(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2019\cdot2020}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(A=1-\frac{1}{2020}=\frac{2019}{2020}< \frac{2020}{2020}=1\)
b) Ta có : \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{1000}}\)
\(2A=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{1001}}\)
\(2A-A=\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{1001}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{1000}}\right)\)
\(A=\frac{1}{2^{1001}}-\frac{1}{2}\)
Tới đây là so sánh đi nhé
22 tháng 7 2020
Cái này mình làm hôm qua rồi mà '-'
a) Đặt \(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2019\cdot2020}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(A=\frac{1}{1}-\frac{1}{2020}=\frac{2019}{2020}\)
\(\Rightarrow A< 1\)
b) \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{1000}}\)
\(2A=2\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{1000}}\right)\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{999}}\)
\(2A-A=A\)
\(=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{999}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{1000}}\right)\)
\(=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^{999}}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{1000}}\)
\(=1-\frac{1}{2^{1000}}\)
\(\Rightarrow A=1-\frac{1}{2^{1000}}< 1\left(đpcm\right)\)
TT
16 tháng 9 2020
Ta có : \(N=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1000.1001}\)
\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{1001-1000}{1000.1001}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1000}-\frac{1}{1001}\)
\(=1-\frac{1}{1001}=\frac{1000}{1001}\)
Ta thấy : \(1001< 2020\Rightarrow\frac{1}{1001}>\frac{1}{2020}\)
\(\Rightarrow-\frac{1}{1001}< -\frac{1}{2020}\)
\(\Rightarrow1-\frac{1}{1001}< 1-\frac{1}{2020}\Rightarrow\frac{1000}{1001}< \frac{2019}{2020}\)
Hay : \(N< M\)
ML
5 tháng 7 2017
a) \(\frac{2^{10}+1}{2^{10}-1}\)và \(\frac{2^{10}-1}{2^{10}-3}\)
Ta có chính chất phân số trung gian là \(\frac{2^{10}+1}{2^{10}-3}\)
\(\frac{2^{10}+1}{2^{10}-1}>\frac{2^{10}+1}{2^{10}-3}\) ; \(\frac{2^{10}-1}{2^{10}-3}< \frac{2^{10}+1}{2^{10}-3}\)
Vì \(\frac{2^{10}+1}{2^{10}-1}>\frac{2^{10}+1}{2^{10}-3}>\frac{2^{10}-1}{2^{10}-3}\)
Nên \(\frac{2^{10}+1}{2^{10}-1}>\frac{2^{10}-1}{2^{10}-3}\)
b) \(A=\frac{2011}{2012}+\frac{2012}{2013}\)và \(B=\frac{2011+2012}{2012+2013}\)
Ta có : \(A=\frac{2011}{2012}+\frac{2012}{2013}>\frac{2011}{2013}+\frac{2012}{2013}=\frac{2011+2012}{2013}>\frac{2011+2012}{2012+2013}=B\)
Vậy A > B
Có gì sai cho sorry
5 tháng 7 2017
a,
\(\frac{2^{10}+1}{2^{10}-1}=1+\frac{2}{2^{10}-1}< 1+\frac{2}{2^{10}-3}=\frac{2^{10}-1}{2^{10}-3}\)
b,
\(\frac{2011}{2012}+\frac{2012}{2013}>\frac{2011}{2012+2013}+\frac{2012}{2012+2013}=\frac{2011+2012}{2012+2013}\)
đặt B=1/2.3+1/3.4+...+1/2011.2012
ta có U =1+1/1.2+1/1.2.3+...+1/1.2.3....2012
B=1/1.2+1/2.3+1/3.4+...+1/2011.2012
=1-1/2+1/2-1/3+...+1/2011-1/2012
=1-1/2012<1 (1)
Mà 1<2(2)
A =1+1/1+1/1.2+1/1.2.3+...+1/1.2.3...2012<1-1/2+1/2-1/3+...+1/2011-1/2012 (3)
từ (1),(2),(3) =>U<2