Dat A=1/5^2+1/6^2+1/7^2+............1/100^2<1/4.5+1/5.6+1/6.7+....+1/99.10=
1/4-1/5+1/5-1/6+1/6-1/7+.............1/99-1/100=
14-1/100=25/100-1/100=24/25/100=1/4(1)
A>1/5.6+1/6.7+1/7.8+....+1/100.101=
1/5-1/6+1/6-1/7+1/7-1/8 +...+1/100-1/101=
1/5-1/101>6 (2)
Tu 1 va 2 => dieu can chung minh
cmr \(\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+......+\frac{1}{100^2}< \frac{1}{4}\)
CMR:\(\frac{1}{6}<\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+.....+\frac{1}{100^2}<\frac{1}{4}\)
giúp mk với các bn . CMR \(\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+..+\frac{1}{100^2}< \frac{1}{4}\)
.CMR: $\frac{1}{3}$ . $\frac{4}{6}$ . $\frac{7}{9}$...$\frac{100}{102}$ < $\frac{1}{17}$
CMR:\(\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\)\(\frac{1}{100^2}< \frac{1}{4}\)
\(cho:A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)
\(CMR:\frac{7}{12}< A< \frac{5}{6}\)
Cho A=\(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}\)
CMR \(\frac{1}{6}< A< \frac{1}{4}\)
CMR:\(\frac{1}{5^3}+\frac{1}{6^3}+\frac{1}{7^3}+...+\frac{1}{2013^3}< \frac{1}{40}\)
cho\(A=\frac{1}{1.2}+\frac{1}{3.4}+.........+\frac{1}{99.100}CMR\frac{7}{12}< A< \frac{5}{6}\)
