\(A=\frac{1}{2}+\frac{1}{12}+...+\frac{1}{9900}>\frac{1}{2}+\frac{1}{12}=\frac{7}{12}\)
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}=\left(1-\frac{1}{2}+\frac{1}{3}\right)-\left(\frac{1}{4}-\frac{1}{5}\right)-...-\left(\frac{1}{98}-\frac{1}{99}\right)-\frac{1}{100}<\left(1-\frac{1}{2}+\frac{1}{3}\right)=\frac{5}{6}\)
Vậy đpcm
cho A=1/1.2+1/3.4+1/5.6+...+1/99.100. chứng minh rằng :7/12 <A<5/16
CHỨNG MINH 7/12<A<5/6 với A=1/1.2+1/3.4+1/5.6+.....+1/99.100
Cho A = 1/1.2+1/3.4+1/5.6+...+1/99.100
a Chứng minh A= 1/51+1/52+1/53+...+1/100
b Chứng minh 7/12<A<5/6
chứng minh rằng:
\(\frac{7}{12}\)<A=\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}
Cho \(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+....\frac{1}{99.100}.\)Chứng minh rằng:
a.\(A=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+....+\frac{1}{100}.\)
b.\(\frac{7}{12}< A< \frac{5}{6}.\)
a=1/1.2+1/3.4+1/5.6+....+1/49.50<1 chứng minh rằng a<1
Chứng minh:
\(\frac{7}{12}< \frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}< \frac{5}{6}\)
cho A=1/1.2+1/3.4+1/5.6+....+1/99.100 chung to rang A>7/12
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}.\)
CM. \(\frac{5}{6}< A< \frac{7}{12}\)
