Cho \(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(B=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{50}\)
\(C=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{48}+\frac{1}{50}\)
Chứng minh rằng A=B-2C
K
Khách
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DD
6 tháng 8 2015
\(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{49.50}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-...-\frac{1}{50}=\left(\frac{1}{1}+\frac{1}{2}+...+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(A=B-2C\left(đpcm\right)\)
23 tháng 8 2019
a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
\(\Rightarrow A< 1\)
23 tháng 8 2019
b) \(B=\frac{1}{3}+\left(\frac{1}{3}\right)^2+...+\left(\frac{1}{3}\right)^{100}\)
\(\Rightarrow3B=1+\frac{1}{3}+...+\left(\frac{1}{3}\right)^{99}\)
\(\Rightarrow3B-B=1-\left(\frac{1}{3}\right)^{100}\)
\(\Rightarrow2B=1-\left(\frac{1}{3}\right)^{100}< 1\)
\(\Rightarrow2B< 1\)
\(\Rightarrow B< \frac{1}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 6 2019
\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{60}\)
\(=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-\left(\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+...+\frac{1}{50}+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\)
2/ \(A=\frac{1}{2}+\frac{1}{12}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(A=\frac{7}{12}+\frac{1}{5.6}+\frac{1}{7.8}+...+\frac{1}{99.100}>\frac{7}{12}\)
Tương tự câu trên ta có: \(A=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(A=\frac{1}{51}+...+\frac{1}{60}+\frac{1}{61}+...+\frac{1}{70}+\frac{1}{71}+...+\frac{1}{80}+\frac{1}{81}+...+\frac{1}{90}+\frac{1}{91}+...+\frac{1}{100}\)
\(A< \frac{1}{50}+...+\frac{1}{50}+\frac{1}{60}+...+\frac{1}{60}+\frac{1}{70}+...+\frac{1}{70}+\frac{1}{80}+...+\frac{1}{80}+\frac{1}{90}+...+\frac{1}{90}\)
\(A< 10.\frac{1}{50}+10.\frac{1}{60}+10.\frac{1}{70}+10.\frac{1}{80}+10.\frac{1}{90}\)
\(A< \frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}< \frac{5}{6}\)
3 tháng 7 2016
\(\frac{1}{1.2}+\frac{1}{3.4}+........+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{49}-\frac{1}{50}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{49}+\frac{1}{50}-2\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{50}\right)\)
\(=1+\frac{1}{2}+......+\frac{1}{50}-\left(1+\frac{1}{2}+....+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+........+\frac{1}{50}\)
\(\Rightarrowđpcm\)
KN
3 tháng 7 2016
ta có:1/1.2+1/3.4+1/5.6+...+1/49.50=>1-1/2+1/3-1/4+1/5-1/6+...+1/49-1/50=>(1+1/3+1/5+1/7+...+1/49)-(1/2+1/4+1/6+...+1/50)=>(1+1/2+1/3+...+1/49+1/50)-(1/2+1/4+1/6+...+1/50).2=>(1+1/2+1/3+...+1/49+1/50) -( 1+1/2+1/3+...+1/25)=>1/26+1/27+1/28+...+1/50=1/26+1/27+1/28+...+1/50hay 1/1.2+1/3.4+1/5.6+...+1/49.50=1/26+1/27+1/28+...+1/50
Bấm mình nha
