1+1+1++1+1+1+1+1+1+1+1+1+1+1=?
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Happy New Year!!!
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\(B\)\(=\) \(\frac{1}{2015}\) + \(\frac{2}{2014}\)\(+\) ... \(+\) \(\frac{2014}{2}\) + \(\frac{2015}{1}\)
\(=\) \(\left(1+\frac{1}{2015}\right)+\left(1+\frac{2}{2014}\right)+...+\left(1+\frac{2014}{2}\right)+\left(\frac{2015}{1}-2014\right)\)
\(=\) \(\frac{2015}{2016}+\frac{2016}{2014}+...+\frac{2016}{2}+\frac{2016}{2016}\)
\(=\)\(2016.\left(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}+...+\frac{1}{3}+\frac{1}{2}\right)\)
\(=\)2016
\(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+...+\dfrac{1}{x}=\dfrac{127}{256}\)
Đặt VT là A
\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{2}{x}\)
\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{2}{x}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{x}\right)=\dfrac{127}{256}\)
\(\Leftrightarrow A=1-\dfrac{1}{x}=\dfrac{127}{256}\)
\(\Leftrightarrow\dfrac{1}{x}=\dfrac{129}{256}\)
\(\Rightarrow x=\dfrac{256}{129}\)
Lời giải:
$2\times A=\frac{2}{1\times 3}+\frac{2}{3\times 5}+\frac{2}{5\times 7}+...+\frac{2}{19\times 21}$
$2\times A=\frac{3-1}{1\times 3}+\frac{5-3}{3\times 5}+\frac{7-5}{5\times 7}+...+\frac{21-19}{19\times 21}$
$=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{19}-\frac{1}{21}$
$=1-\frac{1}{21}=\frac{20}{21}$
$\Rightarrow A=\frac{20}{21}: 2= \frac{10}{21}$
=14 , cho tui nha
=14
HAPYY NEW YEAR!!!