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\(S=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}\)
\(S>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(S>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(S>\frac{1}{2}-\frac{1}{10}\)
\(S>\frac{4}{10}=\frac{2}{5}\)
\(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\)
\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{9.10}< S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3\cdot4}+...+\frac{1}{8.9}\)
=>\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}< S< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{8}-\frac{1}{9}\)
=> \(\frac{1}{2}-\frac{1}{10}< S< 1-\frac{1}{9}\)
=> \(\frac{2}{5}< S< \frac{8}{9}\)(dpcm )
\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{49.50}< 1\)
\(S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{49}-\frac{1}{50}< 1\)
\(S=1-\frac{1}{50}< 1\)
\(S=\frac{49}{50}< 1\left(đpcm\right)\)
VÌ \(\frac{1}{2^2}=\frac{1}{2\cdot2}< \frac{1}{1\cdot2};\frac{1}{3^2}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3};...........;\frac{1}{99^2}=\frac{1}{99\cdot99}< \frac{1}{99\cdot100}\)
\(\Rightarrow S< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+.....+\frac{1}{99\cdot100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\)\(=1-\frac{1}{100}< 1\)\(\Rightarrow S< 1\)
VÌ \(\frac{1}{2\cdot3}< \frac{1}{2\cdot2};.....;\frac{1}{98\cdot99}< \frac{1}{99\cdot99}\)
\(\Rightarrow\)\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+......+\frac{1}{98\cdot99}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{100}=\frac{50}{100}-\frac{1}{100}=\frac{49}{100}< S\)
\(\Rightarrow\frac{49}{100}< S< 1\)
\(K\)\(mk\)\(nha\)
\(a)\) Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\) ta có :
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A< 1-\frac{1}{100}=\frac{99}{100}< 1\)
Vậy \(A< 1\)
Chúc bạn học tốt ~
2S = 1 + 1/2 + 1/2^2 + ... + 1/2^29
2S - S = 1- 1/2^29
S = 1 - 1/2^29 < 1
vậy S < 1
S=\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{30}}\)
2S= \(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{29}}\)
2S - S=( \(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{29}}\)) - (\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{30}}\))
S= \(1-\frac{1}{2^{30}}\)
S= \(\frac{2^{30}}{2^{30}}-\frac{1}{2^{30}}\)
S= \(\frac{2^{30}-1}{2^{30}}\)
Ta có:
S = 1/22 + 1/32 + 1/42 + ... + 1/20162
= 1/2.2 + 1/3.3 + 1/4.4 + ... + 1/2016.2016
S < 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/2015.2016
S < 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/2015 - 1/2016
S < 1 - 1/2016
Mà 1 - 1/2016 < 1
=> S < 1
Vậy S < 1
Ủng hộ nha
\(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2018}}\)
\(2S=2.\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2018}}\right)\)
\(2S=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2017}}\)
\(2S-S=\left(1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2017}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^{2018}}\right)\)
\(S=1-\frac{1}{2^{2018}}< 1\)
\(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{...1}{2^{2018}}\)
\(\Rightarrow2S=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2017}}\)
\(2S-S=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2017}}-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{...1}{2^{2018}}\right)\)
\(S=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2017}}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{2018}}\)
\(S=1-\frac{1}{2^{2018}}\)
\(Mà 1-\frac{1}{2^{2018}}< 1\)
\(\Rightarrow S< 1\)