\(S=3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\)

\(2S=6+3+\frac{3}{2}+...+\frac{3}{2^8}\)

\(2S-S=\left(6+3+\frac{3}{2}+...+\frac{3}{2^8}\right)-\left(3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\right)\)

\(S=6-\frac{3}{2^9}=6-\frac{3}{512}=\frac{3069}{512}\)

\(S=3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\)

\(\Leftrightarrow S=3.\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)

- 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 \(3.D=3.\left(\frac{\frac{1}{2^{10}}-1}{-\frac{1}{2}}\right)=3.\frac{1023}{512}=\frac{3069}{512}\)

Ta có: \(\frac{1}{2}S=\frac{1}{2}.\left(3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\right)\))

=\(\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^{10}}\)

=> \(S-\frac{1}{2}S=\left(3+\frac{3}{2}+...+\frac{3}{2^9}\right)-\left(\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^{10}}\right)\)

=> \(\frac{1}{2}S=3-\frac{3}{2^{10}}\)

=>\(S=\left(3-\frac{3}{2^{10}}\right).2=6-\frac{6}{2^{10}}=6-\frac{3}{2^9}\)

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

\(S=3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\)

\(\Rightarrow S=3.\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)

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 !!! 

Thanks bạn

a, Ta có: \(\frac{1}{2^2}>\frac{1}{2.3};\frac{1}{3^2}>\frac{1}{3.4};...;\frac{1}{10^2}>\frac{1}{10.11}\)

\(\Rightarrow S>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}=\frac{1}{2}-\frac{1}{11}=\frac{9}{22}\)

Vậy S > 9/22

b, Ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{10^2}< \frac{1}{9.10}\)

\(\Rightarrow S>\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}=1-\frac{1}{10}=\frac{9}{10}\)

Vậy S > 9/10

S = \(3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\)

\(=3\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)

Đặt A = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)

2A = \(2\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)

= \(2+1+\frac{1}{2}+....+\frac{1}{2^8}\)

\(2A-A=\left(2+1+\frac{1}{2}+...+\frac{1}{2^8}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)

\(A=2-\frac{1}{2^9}\)

\(\Rightarrow S=3\left(2-\frac{1}{2^9}\right)=\frac{3.\left(2^{10}-1\right)}{2^9}\)

S = 0.5397677312

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