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\(\text{đặt }A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{2016^2}\)
\(A=\frac{1}{2^2}.\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{1008^2}\right)\)
\(A< \frac{1}{2^2}.\left(\frac{1}{1}+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1007.1008}\right)\)
\(A< \frac{1}{4}.\left(1+1-\frac{1}{2007}\right)< \frac{1}{4}.2=\frac{1}{2}\Rightarrow A< \frac{1}{2}\left(ĐPCM\right)\)
\(\dfrac{1}{25}\)S=1/54-1/56+1/58-1/510+...+1/52012-1/52014
\(\Rightarrow\)26/25.S=1/52+1/52014=1/26+...>1/26
đề có lộn không em, chị không biết giải như vậy có đúng đề không
\(S=\left(\frac{1}{2^2}+\frac{1}{2^6}+...+\frac{1}{2^{4n-2}}+..+\frac{1}{2^{2002}}\right)-\left(\frac{1}{2^4}+\frac{1}{2^8}+..+\frac{1}{2^{4n}}+...+\frac{1}{2^{2004}}\right)\)= A - B
Tính A:
\(2^4.A=2^2+\frac{1}{2^2}+\frac{1}{2^6}+...+\frac{1}{2^{4n-2}}+...+\frac{1}{2^{1998}}\)
=> 24.A - A = 15.A =
\(\left(2^2+\frac{1}{2^2}+\frac{1}{2^6}+...+\frac{1}{2^{4n-2}}+...+\frac{1}{2^{1998}}\right)\)- \(\left(\frac{1}{2^2}+\frac{1}{2^6}+...+\frac{1}{2^{4n-2}}+...+\frac{1}{2^{2002}}\right)\)
= 22 - \(\frac{1}{2^{2002}}\) => A = \(\frac{2^2}{15}-\frac{1}{15.2^{2002}}<\frac{4}{15}\)
Tính B :
\(2^4.B=1+\frac{1}{2^4}+\frac{1}{2^8}+...+\frac{1}{2^{4n}}+...+\frac{1}{2^{2000}}\)
=> 24.B - B
=\(\left(1+\frac{1}{2^4}+\frac{1}{2^8}+...+\frac{1}{2^{4n}}+...+\frac{1}{2^{2000}}\right)\)- \(\left(\frac{1}{2^4}+\frac{1}{2^8}+...+\frac{1}{2^{4n}}+...+\frac{1}{2^{2004}}\right)\)
= \(1-\frac{1}{2^{2004}}\Rightarrow B=\frac{1}{15}-\frac{1}{15.2^{2004}}<\frac{1}{15}\)
Vậy S < \(\frac{4}{15}-\frac{1}{15}=\frac{3}{15}=\frac{1}{5}=0,2\) ĐPCM
Có S=\(\dfrac{1}{2^2}-\dfrac{1}{2^4}+\dfrac{1}{2^6}-...+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+...+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\)
=>\(\dfrac{1}{2^2}S=\dfrac{1}{2^2}\)\(\left(\dfrac{1}{2^2}-\dfrac{1}{2^4}+\dfrac{1}{2^6}-...+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+...+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\right)\)
=> \(\dfrac{1}{2^2}\)S= \(\dfrac{1}{2^4}-\dfrac{1}{2^6}+\dfrac{1}{2^8}-...+\dfrac{1}{2^{4n}}-\dfrac{1}{2^{4n+2}}+...+\dfrac{1}{2^{2004}}-\dfrac{1}{2^{2006}}\)
+S =\(\dfrac{1}{2^2}-\dfrac{1}{2^4}+\dfrac{1}{2^6}-...+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+...+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\)
=> \(\dfrac{5}{4}\)S= \(\dfrac{1}{2^2}\)-\(\dfrac{1}{2^{2006}}\)
=> S= \(\dfrac{\left(\dfrac{1}{2^2}-\dfrac{1}{2^{2006}}\right)}{\dfrac{5}{2^2}}=\dfrac{\dfrac{1}{2^2}}{\dfrac{5}{2^2}}-\dfrac{\dfrac{1}{2^{2006}}}{\dfrac{5}{2^2}}=\dfrac{1}{5}-\dfrac{1}{2^{2004}.5}=0.2-\dfrac{1}{2^{2004}.5}\)
=> S <0,2
Vậy S <0,2(đpc/m)
Nếu 1/2^2*S=1/2^2 thì tính đc S luôn r cần gì làm nữa bạn
Cũng cảm ơn vì đã giúp nhé
2
a) (2x - 1)4 = 81
<=> \(\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=4\\2x=-2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}}\)
a)ta có 3B=1+1/3+1/3^2+........+1/3^2003+1/3^2004
B= 1/3+1/3^2+........+1/3^2003+1/3^2004+1/3^2005
suy ra 2B=1-1/3^2005
suy ra B=\(\frac{1-\frac{1}{3}^{2005}}{2}\)
suy ra B=1/2-1/3^2005/2 bé hơn 1/2
từ đấy suy ra B bé hơn 1/2