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ta có
S = \(\frac{1}{2}\)+ \(\frac{1}{2^2}\)+ \(\frac{1}{2^3}\)+ .....+\(\frac{1}{20^{20}}\)
2S= 1 + \(\frac{1}{2}\)+ \(\frac{1}{2^2}\)+ \(\frac{1}{2^3}\)+ .....+\(\frac{1}{2^{19}}\)
S = 2S-S= 1 - \(\frac{1}{2^{19}}\)<1
Vậy S < 1
^_^ chúc bn học tốt
a)Ta có: \(\frac{3}{1.4}=\frac{4-1}{1.4}=1-\frac{1}{4}\)
\(\frac{3}{4.7}=\frac{7-4}{4.7}=\frac{1}{4}-\frac{1}{7}\)
... . . . .
\(\frac{3}{n\left(n+3\right)}=\frac{1}{n}-\frac{1}{n+3}\)
\(\Leftrightarrow S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n+3}< 1^{\left(đpcm\right)}\)
b) Ta có: \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
Suy ra \(\frac{2}{5}< S\) (1)
Ta lại có: \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
Mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}=1-\frac{1}{9}=\frac{8}{9}\)
Từ đó suy ra S < 8/9
Từ (1) và (2) suy ra đpcm
\(\Rightarrow2S=1+\frac{1}{2}+...+\frac{1}{2^{19}}\)
\(\Rightarrow2S-S=\left(1+\frac{1}{2}+...+\frac{1}{2^{19}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{20}}\right)\)
\(\Rightarrow S=1-\frac{1}{2^{20}}< 1\)
Ta có: S = \(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{20}}\)
1/2S = \(\frac{1}{2}.\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{20}}\right)\)
1/2S = \(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{21}}\)
1/2S - S = \(\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{21}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{20}}\right)\)
-1/2S = \(\frac{1}{2^{21}}-\frac{1}{2}\)
S = \(\left(\frac{1}{2^{21}}-\frac{1}{2}\right):\left(-\frac{1}{2}\right)\)
S =\(\frac{1}{2^{21}}:\left(-\frac{1}{2}\right)-\frac{1}{2}:\left(-\frac{1}{2}\right)\)
S = \(-\frac{1}{2^{20}}+1=1-\frac{1}{2^{20}}< 1\)
Ta có : S =\(\frac{1}{2^2}\)\(+\)\(\frac{1}{3^2}\)\(+\)...\(+\)\(\frac{1}{9^2}\)
= \(\frac{1}{2.2}\)\(+\)\(\frac{1}{3.3}\)\(+\)...\(+\)\(\frac{1}{9^2}\)
\(\Rightarrow\)S > \(\frac{1}{2.3}\)\(+\)\(\frac{1}{3.4}\)\(+\)...\(+\)\(\frac{1}{9.10}\)
= \(\frac{1}{2}\)\(-\)\(\frac{1}{3}\)\(+\)\(\frac{1}{3}\)\(-\)\(\frac{1}{4}\)\(+\)..\(+\)\(\frac{1}{9}\)\(-\)\(\frac{1}{10}\)
= \(\frac{1}{2}\)\(-\)\(\frac{1}{10}\)
\(\Rightarrow\)S < \(\frac{1}{1.2}\)\(+\)\(\frac{1}{2.3}\)\(+\)...\(+\)\(\frac{1}{8.9}\)
=\(1\)\(-\)\(\frac{1}{2}\)\(+\)\(\frac{1}{2}\)\(-\)\(\frac{1}{3}\)\(+\)...\(+\)\(\frac{1}{8}\)\(-\)\(\frac{1}{9}\)
= \(1\)\(-\)\(\frac{1}{9}\)= \(\frac{8}{9}\)
Vậy \(\frac{2}{5}\)< S < \(\frac{8}{9}\)(đpcm)
Chúc bạn học tốt
a)\(S=\left(\frac{1}{5}+\frac{1}{13}\right)+\left(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\right)<\left(\frac{1}{5}+\frac{1}{5}\right)+\left(\frac{1}{60}+\frac{1}{60}+\frac{1}{60}\right)=\frac{2}{5}+\frac{1}{20}=\frac{9}{20}<\frac{10}{20}=\frac{1}{2}\)b) \(2.S=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{19}}\)
=> 2S - S = \(\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{19}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{20}}\right)\)
=> S = \(1-\frac{1}{2^{20}}<1\) đpcm
\(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}...+\frac{1}{2^{20}}\)
\(2S=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{19}}\)
Ta lấy \(2S-S\)được
\(1-\frac{1}{2^{19}}\)
\(\Rightarrow S=1-\frac{1}{2^{19}}< 1\left(ĐPCM\right)\)
\(S=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{20}}< 1\)
\(2S=1+\frac{1}{2}+...+\frac{1}{2^{19}}\)
\(=>2S-S=1-\frac{1}{2^{19}}\)
\(=>S=1-\frac{1}{2^{19}}< 1\left(đpcm\right)\)