Cho S= 1/2 + 1/8 + 1/18 + 1/32 + 1/50 + 1/72 + 1/98 + 1/128 + 1/162
Chứng tỏ S < 18/19
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Tổng S có 50 phân số
=> S > 1/100 + 1/100 + 1/100 +...+ 1/100 (50 phân số) => S > 1/2.
Vậy S > 1/2
Tổng S có 50 phân số
=> S > 1/100 + 1/100 + 1/100 +...+ 1/100 (50 phân số) => S > 1/2.
Vậy S > 1/2
\(S=\left(\frac{1}{50}+\frac{1}{51}+...+\frac{1}{74}\right)+\left(\frac{1}{75}+\frac{1}{76}+...+\frac{1}{99}\right)\)
Có: \(\frac{1}{50}+\frac{1}{51}+...+\frac{1}{74}>\frac{1}{75}+\frac{1}{75}+...+\frac{1}{75}=\frac{25}{75}=\frac{1}{3}\)
\(\frac{1}{75}+\frac{1}{76}+...+\frac{1}{99}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{25}{100}=\frac{1}{4}\)
=> \(S>\frac{1}{3}+\frac{1}{4}=\frac{7}{12}>\frac{6}{12}=\frac{1}{2}\)=> đpcm
Giải:
\(S=\dfrac{1}{50}+\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{98}+\dfrac{1}{99}\)
\(S=\left(\dfrac{1}{50}+\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{74}\right)+\left(\dfrac{1}{75}+...+\dfrac{1}{98}+\dfrac{1}{99}\right)\)
\(\Rightarrow S>\left(\dfrac{1}{50}+\dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}\right)+\left(\dfrac{1}{75}+...+\dfrac{1}{75}+\dfrac{1}{75}\right)\)
\(\Rightarrow S>\dfrac{1}{2}+\dfrac{1}{3}>\dfrac{1}{2}\)
\(\Rightarrow S>\dfrac{1}{2}\left(đpcm\right)\)
\(S=\dfrac{1}{50}+\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{98}+\dfrac{1}{99}\)
\(S=\dfrac{1}{50}>100\) \(\dfrac{1}{51}>100\) \(\dfrac{1}{52}>100\) \(....\) \(\dfrac{1}{98}>100\) \(\dfrac{1}{99}>100\)
\(\Rightarrow S>\dfrac{1}{100}+\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}+\dfrac{1}{100}\\ \) {50 số 100}
\(S>50\cdot\dfrac{1}{100}=\dfrac{1}{2}\)
\(S>\dfrac{1}{2}\)
b: A=1/3+1/9+...+1/3^10
=>3A=1+1/3+...+1/3^9
=>A*2=1-1/3^10=(3^10-1)/3^10
=>A=(3^10-1)/(2*3^10)
c: C=3/2+3/8+3/32+3/128+3/512
=>4C=6+3/2+...+3/128
=>3C=6-3/512
=>C=1023/512
d: A=1/2+...+1/256
=>2A=1+1/2+...+1/128
=>A=1-1/256=255/256