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\(A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2014}}\)
\(\Rightarrow3A=3+1+\frac{1}{3}+...+\frac{1}{3^{2013}}\)
\(\Rightarrow3A-A\)= \(\left(3+1+...+\frac{1}{3^{2013}}\right)-\left(1+\frac{1}{3}+...+\frac{1}{3^{2014}}\right)\)
\(\Rightarrow2A=3-\frac{1}{3^{2014}}\)
\(\Rightarrow A=\frac{3-\frac{1}{3^{2014}}}{2}\)
\(\Rightarrow A=\frac{3}{2}-\frac{\frac{1}{3^{2014}}}{2}< \frac{3}{2}\)
Vậy \(A< \frac{3}{2}\)
Chúc bạn học tốt !!!
Mấy bài dạng này biết cách làm là oke
Ta có :
\(A=\frac{\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=\frac{\left(2016-1-1-...-1\right)+\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{2}{2015}+1\right)+\left(\frac{1}{2016}+1\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=\frac{\frac{2017}{2017}+\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2015}+\frac{2017}{2016}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=\frac{2017\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=2017\)
Vậy \(A=2017\)
Chúc bạn học tốt ~
\(A=\frac{\frac{2016}{1}+\frac{2015}{2}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=\frac{2016+\frac{2015}{2}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=\frac{\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{2}{2015}+1\right)+\left(\frac{1}{2016}+1\right)+\frac{2017}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
(số 2016 tách ra làm 2016 số 1 rồi cộng vào từng phân số, còn dư 1 số viết thành 2017/2017 nghe bạn!!! :)))
\(A=\frac{\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2015}+\frac{2017}{2016}+\frac{2017}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=\frac{2017\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=2017\)
\(S=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{2014}{5^{2014}}\)
\(5S=1+\frac{2}{5}+\frac{3}{5^2}+...+\frac{2014}{5^{2013}}\)
\(\Rightarrow5S-S=1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2013}}-\frac{2014}{5^{2014}}\)
\(S=\frac{1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2013}}-\frac{2014}{5^{2014}}}{4}\)
Xét \(A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2013}}\)
\(5A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2012}}\)
\(5A-A=1-\frac{1}{5^{2013}}\Leftrightarrow A=\frac{1-\frac{1}{5^{2013}}}{4}=\frac{1}{4}-\frac{1}{4.5^{2013}}\)
\(\Rightarrow S=\frac{1+\frac{1}{4}-\left(\frac{1}{4.5^{2013}}+\frac{2014}{5^{2014}}\right)}{4}=\frac{5}{16}-\frac{\frac{1}{4.5^{2013}}+\frac{2014}{5^{2014}}}{4}< \frac{1}{3}\)
\( P=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2014^2}< Q= 1\frac{3}{4}\)
Ta có
\(\frac{A}{B}=\frac{1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{4026}}{1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{4025}}\)
\(\Rightarrow\frac{A}{B}=\frac{\left(1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{4025}\right)+\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{4026}\right)}{1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{4025}}\)
\(\Rightarrow\frac{A}{B}=\frac{1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{4025}}{1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{4025}}+\frac{\frac{1}{2}+\frac{1}{4}+....+\frac{1}{4026}}{1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{4025}}\)
\(\Rightarrow\frac{A}{B}=1+\frac{\frac{1}{2}+\frac{1}{4}+....+\frac{1}{4026}}{1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{4025}}\)
Dễ thấy A/B > 1
2013/2014<1
=> \(\frac{A}{B}>\frac{2013}{2014}\)
\(1\dfrac{2013}{2014}\) cơ mà sao lại \(\dfrac{2013}{2014}\)
\(3A=3\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2014}}\right)\)
\(3A=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2013}}\)
\(3A-A=\left(3+1+\frac{1}{3}+...+\frac{1}{3^{2013}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2014}}\right)\)
\(2A=3-\frac{1}{3^{2014}}\)
\(A=\left(3-\frac{1}{3^{2014}}\right):2\)
\(A=\frac{3}{2}-\frac{1}{2.3^{2014}}<\frac{3}{2}\)
\(\Rightarrow A<\frac{3}{2}\)