Cho M = \(\dfrac{3}{4}+\dfrac{3^2}{4^2}+\dfrac{3^3}{4^3}+...+\dfrac{3^{2016}}{4^{2016}}\). Tìm phần nguyên của M.
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Akai Haruma
Giáo viên
8 tháng 7 2017
Lời giải:
Có \(M=\left ( \frac{1}{4}+\frac{3}{4^3}+...+\frac{2015}{4^{2015}} \right )-\left ( \frac{2}{4^2}+\frac{4}{4^4}+...+\frac{2016}{4^{2016}} \right )=A-B\)
Xét \(A= \frac{1}{4}+\frac{3}{4^3}+...+\frac{2015}{4^{2015}} \Rightarrow 16A=4+\frac{3}{4}+\frac{5}{4^3}+...+\frac{2015}{4^{2013}}\)
\(\Rightarrow 15A=4+2\underbrace{\left ( \frac{1}{4}+\frac{1}{4^3}+...+\frac{1}{4^{2013}} \right )}_{T}-\frac{2015}{4^{2015}}\)
Lại có \(16T=4+\frac{1}{4}+\frac{1}{4^3}+...+\frac{1}{4^{2011}}\Rightarrow 15T=4-\frac{1}{4^{2013}}\)
Do đó \(A=\frac{1}{15}\left ( 4+\frac{8}{15}-\frac{2}{15.4^{2013}}-\frac{2015}{4^{2015}} \right )\)
Thực hiện tương tự, suy ra
\(B=\frac{1}{15}\left ( 2+\frac{2}{15}-\frac{2}{15.4^{2014}}-\frac{2016}{4^{2016}} \right )\)
\(\Rightarrow M=A-B=\frac{1}{15}\left ( \frac{12}{5}-\frac{90692}{15.4^{2014}} \right )<\frac{1}{15}.\frac{12}{5}=\frac{4}{25}\)
Ta có đpcm
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14 tháng 12 2017
\(B=\dfrac{2016}{1}+\dfrac{2015}{2}+\dfrac{2014}{3}+...+\dfrac{3}{2014}+\dfrac{2}{2015}+\dfrac{1}{2016}\)
\(B=2016+\dfrac{2015}{2}+\dfrac{2014}{3}+....+\dfrac{3}{2014}+\dfrac{2}{2015}+\dfrac{1}{2016}\)
\(B=1+\left(\dfrac{2015}{2}+1\right)+\left(\dfrac{2014}{3}+1\right)+...+\left(\dfrac{3}{2014}+1\right)+\left(\dfrac{2}{2015}+1\right)+\left(\dfrac{1}{2016}+1\right)\)
\(B=\dfrac{2017}{2017}+\dfrac{2017}{2}+\dfrac{2017}{3}+....+\dfrac{2017}{2014}+\dfrac{2017}{2015}+\dfrac{2017}{2016}\)
\(B=2017\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)\)
\(\dfrac{B}{A}=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}}=2017\)
14 tháng 12 2017
\(\dfrac{B}{A}=\dfrac{\dfrac{2016}{1}+\dfrac{2015}{2}+\dfrac{2014}{3}+...+\dfrac{3}{2014}+\dfrac{2}{2015}+\dfrac{1}{2016}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=\dfrac{1+\left(\dfrac{2015}{2}+1\right)+\left(\dfrac{2014}{3}+1\right)+...+\left(\dfrac{2}{2015}+1\right)+\left(\dfrac{1}{2016}+1\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=\dfrac{\dfrac{2017}{2017}+\left(\dfrac{2015}{2}+\dfrac{2}{2}\right)+\left(\dfrac{2014}{3}+\dfrac{3}{3}\right)+...+\left(\dfrac{1}{2016}+\dfrac{2016}{2016}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=2017\)
Vậy \(\dfrac{B}{A}=2017\)
Lời giải:
\(M=\frac{3}{4}+\frac{3^2}{4^2}+\frac{3^3}{4^3}+..+\frac{3^{2016}}{4^{2016}}\)
\(\Rightarrow \frac{4}{3}M=1+\frac{3}{4}+\frac{3^2}{4^2}+..+\frac{3^{2015}}{4^{2015}}\)
Trừ theo vế:
\(\frac{4}{3}M-M=1-\frac{3^{2016}}{4^{2016}}\)
\(\Rightarrow M=3-\frac{3^{2017}}{4^{2016}}\)
\(\Rightarrow \left \lfloor M \right \rfloor=\left \lfloor 3-\frac{3^{2017}}{4^{2016}} \right \rfloor\)
Ta thấy \(2<3-\frac{3^{2017}}{4^{2016}}<3\) nên \(\Rightarrow \left \lfloor M \right \rfloor=\left \lfloor 3-\frac{3^{2017}}{4^{2016}} \right \rfloor=2\)
Sao \(2< 3-\dfrac{3^{2017}}{4^{2016}}\) được vậy ?