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a, Ta có :\(A=\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{49}}+\dfrac{1}{2^{50}}\\ \Rightarrow2A=1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{49}}\\ \Rightarrow2A-A=\left(1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{49}}\right)-\left(\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{50}}\right)\\ \Rightarrow A=1-\dfrac{1}{2^{50}}< 1\\ \Rightarrow A< 1\) Vậy \(A< 1\)
b, Ta có :
\(B=\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\\ \Rightarrow3B=1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\\ \Rightarrow3B-B=\left(1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\right)\\ \Rightarrow2B=1-\dfrac{1}{3^{100}}< 1\\ \Rightarrow B< \dfrac{1}{2}\)Vậy \(B< \dfrac{1}{2}\)
c, Ta có :
\(C=\dfrac{1}{4^1}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{1000}}\\ \Rightarrow4C=1+\dfrac{1}{4^1}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{999}}\\\Rightarrow4C-C=\left(1+\dfrac{1}{4^1}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{999}}\right)-\left(\dfrac{1}{4^1}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{1000}}\right)\\ \Rightarrow3C=1-\dfrac{1}{4^{1000}}< 1\\ \Rightarrow C< \dfrac{1}{3}\)Vậy \(C< \dfrac{1}{3}\)
a/ Đặt :
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+.........+\dfrac{1}{3^{50}}\)
\(\Leftrightarrow3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+.......+\dfrac{1}{3^{49}}\)
\(\Leftrightarrow3A-A=\left(1+\dfrac{1}{3}+....+\dfrac{1}{3^{49}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+....+\dfrac{1}{3^{50}}\right)\)
\(\Leftrightarrow2A=1-\dfrac{1}{3^{50}}\)
còn sao nx thì mk chịu =.=
\(S=\dfrac{2}{2^1}+\dfrac{3}{2^2}+...+\dfrac{2017}{2^{2016}}\)
\(\Rightarrow2S=2+\dfrac{3}{2^1}+\dfrac{4}{2^2}+...+\dfrac{2017}{2^{2015}}\)
\(\Rightarrow2S-S=\left(2+\dfrac{3}{2^1}+\dfrac{4}{2^2}+...+\dfrac{2017}{2^{2015}}\right)-\left(\dfrac{2}{2^1}+\dfrac{3}{2^2}+...+\dfrac{2017}{2^{2016}}\right)\)
\(\Leftrightarrow S=2+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2015}}-\dfrac{2017}{2^{2016}}\)
Tới đây thì đơn giản rồi nhé
Giải:
Ta có:
\(\dfrac{A}{B}=\dfrac{1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{4026}}{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{4025}}\)
\(\Rightarrow\dfrac{A}{B}=\dfrac{\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{4025}\right)+\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2046}\right)}{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{4025}}\)
\(\Rightarrow\dfrac{A}{B}=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{4025}}{1+\dfrac{1}{3}+\dfrac{1}{5}+....+\dfrac{1}{4025}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{4026}}{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{4025}}\)
\(\Rightarrow\dfrac{A}{B}=1+\dfrac{\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2046}}{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{4025}}\)
Dễ thấy \(\dfrac{A}{B}>1\)
Mà \(\dfrac{2013}{2014}< 1\)
\(\Rightarrow\dfrac{A}{B}>1\dfrac{2013}{2014}\)
Ta thấy:
\(1\cdot2^2=2^2;2\cdot3^2>3^2;3\cdot4^2>4^2;...;49\cdot50^2>50^2\)
\(\Rightarrow\dfrac{1}{1.2^2}=\dfrac{1}{2^2};\dfrac{1}{2\cdot3^2}< \dfrac{1}{3^2};\dfrac{1}{3\cdot4^2}< \dfrac{1}{4^2};...;\dfrac{1}{49\cdot50^2}< \dfrac{1}{50^2}\)
\(\Rightarrow\dfrac{1}{1\cdot2^2}+\dfrac{1}{2\cdot3^2}+\dfrac{1}{3\cdot4^2}+...+\dfrac{1}{49\cdot50^2}< \dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\)
hay A<B
Vậy A<B
B=\(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2012}}\)
=>3B=\(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2011}}\)
=>3B-B=2B=1-\(\dfrac{1}{3^{2012}}\)
=>B=\(\dfrac{1}{2}-\dfrac{1}{2.3^{20112}}\)<1/2
vậy........
\(A=\left(\dfrac{1}{2}-\dfrac{7}{13}-\dfrac{1}{3}\right)+\left(\dfrac{-6}{13}+\dfrac{1}{2}+1\dfrac{1}{3}\right)\)
\(A=\dfrac{1}{2}-\dfrac{7}{13}-\dfrac{1}{3}-\dfrac{6}{13}+\dfrac{1}{2}+\dfrac{4}{3}\)
\(A=\left(\dfrac{1}{2}+\dfrac{1}{2}\right)-\left(\dfrac{7}{13}+\dfrac{6}{13}\right)+\left(\dfrac{4}{3}-\dfrac{1}{3}\right)\)
\(A=1-1+1=1\)
\(B=\left(-1\dfrac{1}{2}:\dfrac{3}{-4}\right).\left(-4\dfrac{1}{2}\right)-\dfrac{1}{4}\)
\(B=\dfrac{-3}{2}:\dfrac{3}{-4}.\dfrac{-9}{2}-\dfrac{1}{4}\)
\(B=2.\dfrac{-9}{2}-\dfrac{1}{4}\)
\(=-9-\dfrac{1}{4}=\dfrac{-37}{4}\)
\(a,A=\left(\dfrac{1}{2}-\dfrac{7}{13}-\dfrac{1}{3}\right)+\left(-\dfrac{6}{13}+\dfrac{1}{2}+1\dfrac{1}{3}\right)\)
\(A=\dfrac{1}{2}-\dfrac{7}{13}-\dfrac{1}{3}+\dfrac{-6}{13}+\dfrac{1}{2}+\dfrac{4}{3}\)
\(A=\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(-\dfrac{7}{13}-\dfrac{6}{13}\right)+\left(-\dfrac{1}{3}+\dfrac{4}{3}\right)\)
\(A=-1+1=0\)
\(b,B=\left(-1\dfrac{1}{2}:\dfrac{3}{-4}\right)\left(-4\dfrac{1}{2}\right)-\dfrac{1}{4}\)
\(B=\left(-\dfrac{3}{2}.\dfrac{-4}{3}\right).\dfrac{-9}{2}-\dfrac{1}{4}\)
\(B=8.\dfrac{-9}{2}-\dfrac{1}{4}\)
\(B=-36-\dfrac{1}{4}\)
B = \(-\dfrac{145}{4}\)
Sau 4 tháng commend ;-; (mình đặt dãy số trên là A nhé)
Ta có: \(\frac{1}{2!}=\left(\frac{2}{2!}-\frac{1}{2!}\right);\frac{2}{3!}=\left(\frac{3}{3!}-\frac{2}{3!}\right)\frac{3}{4!}=\left(\frac{4}{4!}-\frac{3}{4!}\right)\)
Nhận thấy: \(\frac{n}{\left(n+1\right)!}=\frac{n+1}{\left(n+1\right)!}-\frac{1}{\left(n+1\right)!}\)
Áp dụng tương tự với câu trên, ta có:
\(A=\frac22-\frac{1}{1\times2}+\frac{3}{3\times2\times1}-\frac{1}{3\times2\times1}+\frac{4}{4\times3\times2\times1}-\frac{1}{4\times3\times2\times1}+\cdots+\frac{2025}{2025\times2024\times2023\times\ldots\times1}-\frac{1}{2025\times2024\times2023\times\ldots\times1}\)
\(A=\frac22-\frac{1}{1\times2}+\frac{1}{2\times1}-\frac{1}{3\times2\times1}+\frac{1}{3\times2\times1}-\frac{1}{4\times2\times1}+\cdots+\frac{2025}{2025\times2024\times2023\times\ldots\times1}-\frac{1}{2025\times2024\times2023\times\ldots\times1}\)
Nên: \(A=1-\frac{1}{2025!}\) Mà \(\frac{1}{2025!}>0\)
Vậy\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2024}{2025!}>1-0=1\)
*bé hơn 1 nhé (mình viết nhầm)