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Mk chỉ làm đc bài 2 thôi!
\(S=3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\)
\(\Rightarrow2S=6+3+\frac{3}{2}+...+\frac{3}{2^8}\)
\(\Rightarrow2S-S=6-\frac{3}{2^9}\)
\(\Rightarrow S=6-\frac{3}{2^9}\)
Chúc bạn học tốt ( sai thì đừng ném đá ) !
Ta có :
A = \(\frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{50^2}\)< \(\frac{1}{1.1}+\frac{1}{1.2}+...+\frac{1}{49.50}\)
A < \(1-1+1-\frac{1}{2}+...+\frac{1}{49}-\frac{1}{50}\)
A < 1 - 1/50 = 49/50 < 2
Vậy A < 2
S= - 32\(\left(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+...+\frac{1}{868}\right)\)
S = - 32\(\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{28.31}\right)\)
S = - 3\(\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{28.31}\right)\)
S = -3\(\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{28}-\frac{1}{31}\right)\)
S = -3 \(\left(1-\frac{1}{31}\right)\)
S = -3\(.\frac{30}{31}\)
S = -90/31
1/3S=-(1/1*4+1/4*7+1/7*10+...+1/28*31)=-(1/1-1/4+1/4-1/7+1/7-1/10+...+1/28-1/31)=-(1/1-1/31)=-30/31
=>S=(-30/31):1/3=-90/31
\(S=3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\)
\(\Leftrightarrow S=3.\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)
- 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: \(\frac{1}{2}S=\frac{1}{2}.\left(3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\right)\))
=\(\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^{10}}\)
=> \(S-\frac{1}{2}S=\left(3+\frac{3}{2}+...+\frac{3}{2^9}\right)-\left(\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^{10}}\right)\)
=> \(\frac{1}{2}S=3-\frac{3}{2^{10}}\)
=>\(S=\left(3-\frac{3}{2^{10}}\right).2=6-\frac{6}{2^{10}}=6-\frac{3}{2^9}\)
Xét TH1 : ( S < 8/9 )
\(\frac{1}{2\cdot2}< \frac{1}{1\cdot2};\frac{1}{3\cdot3}< \frac{1}{2\cdot3};...;\frac{1}{9\cdot9}< \frac{1}{8\cdot9}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{8\cdot9}\)
hay \(S< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{8\cdot9}\)
\(S< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}\)
\(S< 1-\frac{1}{9}=\frac{8}{9}\left(1\right)\)
TH2 : ( S > 2/5 )
\(\frac{1}{2\cdot2}>\frac{1}{2\cdot3};\frac{1}{3\cdot3}>\frac{1}{3\cdot4};...;\frac{1}{9\cdot9}>\frac{1}{9\cdot10}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}>\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\)
hay \(S>\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\)
\(S>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(S>\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\left(2\right)\)
Từ (1) và (2) => đpcm
Ko tk thì ko phải là ng` nx rồi :)
S=3/2^0+3/2^1+....+3/2^2018
S=3/2.(2/2^0+2/2^1+....+2^2018)
đặt B=2/2^0+2/2^1+....+2^2018
2B=2.(2/2^0+2/2^1+....+2^2018)
2B=1+2/2^0+...+2/2^2017
2B-B=(1+2/2^0+...+2/2^2017)-(2/2^0+2/2^1+....+2^2018)
B=1-2^2018
S=3/2.1-2^2018=3/2^2018
\(S=3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\)
\(\Rightarrow S=3.\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)
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 !!!
Ta có :
\(S=3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\)
\(2S=6+3+\frac{3}{2}+...+\frac{3}{2^8}\)
\(2S-S=\left(6+3+\frac{3}{2}+...+\frac{3}{2^8}\right)-\left(3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\right)\)
\(S=6-\frac{3}{2^9}\)
\(S=\frac{2^{10}.3-3}{2^9}\)
Vậy \(S=\frac{2^{10}.3-3}{2^9}\)
vận dụng 3S lên
xong tìm S nha bn ok
tại k có thời gian nên chỉ giúp thế thôi
3069/512