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\(S=3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\)
\(\Leftrightarrow S=3.\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)
- Đặt \(D=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
\(\Leftrightarrow\frac{1}{2}D=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
\(\Leftrightarrow\frac{1}{2}D-D=\frac{1}{2^{10}}-1\)
\(\Leftrightarrow D=\frac{\frac{1}{2^{10}}-1}{-\frac{1}{2}}\)
Vậy \(3.D=3.\left(\frac{\frac{1}{2^{10}}-1}{-\frac{1}{2}}\right)=3.\frac{1023}{512}=\frac{3069}{512}\)
\(S=3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\)
\(\Rightarrow S=3.\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)
Đặt \(A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
\(\Rightarrow2A=2+1+\frac{1}{2}+...+\frac{1}{2^8}\)
\(\Rightarrow2A-A=\left(2+1+\frac{1}{2}+...+\frac{1}{2^8}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)
\(\Rightarrow A=2-\frac{1}{2^9}\)
Mà \(S=3.A\)
\(\Rightarrow S=3.\left(2-\frac{1}{2^9}\right)\)
\(\Rightarrow S=6-\frac{3}{2^9}\)
Chúc bạn học tốt !!!
\(S=3+\frac{3}{2}+\frac{3}{2^2}+....+\frac{3}{2^9}\)
\(\Rightarrow2S=6+3+\frac{3}{2}+....+\frac{3}{2^8}\)
\(\Rightarrow2S-S=\left(6+3+\frac{3}{2}+...+\frac{3}{2^8}\right)-\left(3+\frac{3}{2}+\frac{3}{2^2}+....+\frac{3}{2^9}\right)\)
\(\Rightarrow S=6-\frac{3}{2^9}=\frac{3069}{512}\)
\(2S=2\left(3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\right)\)
\(2S=6+3+\frac{3}{2}+...+\frac{3}{2^8}\)
\(2S-S=\left(6+3+\frac{3}{2}+...+\frac{3}{2^8}\right)-\left(3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\right)\)
\(S=6-\frac{3}{2^9}\)
tới đây thì bn tự tính nhé
a )
\(\frac{-4}{9}.\frac{1}{3}-\frac{4}{9}.\frac{5}{6}+\frac{3}{7}.\frac{4}{9}\)
\(=\frac{4}{9}.\left(-\frac{1}{3}-\frac{5}{6}+\frac{3}{7}\right)\)
\(=\frac{4}{9}.\left(-\frac{14}{42}-\frac{35}{42}+\frac{18}{42}\right)\)
\(=\frac{4}{9}.\frac{-31}{42}\)
\(=-\frac{62}{189}\)
b )
\(\frac{2}{3}:\frac{3}{7}-\frac{2}{3}:\frac{4}{3}+\frac{2}{3}:\frac{1}{21}\)
\(=\frac{2}{3}.\frac{7}{3}-\frac{2}{3}.\frac{3}{4}+\frac{2}{3}.21\)
\(=\frac{14}{9}-\frac{1}{2}+14\)
\(=\frac{28}{18}-\frac{9}{18}+14\)
\(=\frac{19}{18}+14\)
\(=1+14+\frac{1}{18}\)
\(=15\frac{1}{18}\)
c )
\(\left(5\frac{1}{3}+3\frac{2}{3}\right)-4\frac{1}{3}\)
\(=\left(5+3-4\right)+\left(\frac{1}{3}+\frac{2}{3}-\frac{1}{3}\right)\)
\(=4\frac{2}{3}\)
\(=\frac{14}{3}\)
a) \(-\frac{4}{9}\cdot\frac{1}{3}-\frac{4}{9}\cdot\frac{5}{6}+\frac{3}{7}\cdot\frac{4}{9}\)
\(=\left(-\frac{4}{9}\right)\cdot\frac{1}{3}+\left(-\frac{4}{9}\right)\cdot\frac{5}{6}-\left(-\frac{4}{9}\right)\cdot\frac{3}{7}\)
\(=\left(-\frac{4}{9}\right)\left(\frac{1}{3}+\frac{5}{6}-\frac{3}{7}\right)\)
\(=\left(-\frac{4}{9}\right)\cdot\frac{31}{42}=-\frac{62}{189}\)
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
\(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)
=>\(2S=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\)
=>\(2S-S=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\right)\)
=>\(S=1-\frac{1}{2^9}=\frac{511}{512}\)
Vậy \(S=\frac{511}{512}\)
Ta có : \(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+......+\frac{1}{2^9}\)
\(\Rightarrow2S=1+\frac{1}{2}+\frac{1}{2^3}+....+\frac{1}{2^8}\)
\(\Rightarrow2S-S=1-\frac{1}{2^9}\)
\(\Leftrightarrow S=1-\frac{1}{2^9}\)
Ta có: \(\frac{1}{2}S=\frac{1}{2}.\left(3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\right)\))
=\(\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^{10}}\)
=> \(S-\frac{1}{2}S=\left(3+\frac{3}{2}+...+\frac{3}{2^9}\right)-\left(\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^{10}}\right)\)
=> \(\frac{1}{2}S=3-\frac{3}{2^{10}}\)
=>\(S=\left(3-\frac{3}{2^{10}}\right).2=6-\frac{6}{2^{10}}=6-\frac{3}{2^9}\)