A=1/2!+1/2!-1/3!+...+1/99!-1/100
=1/2-1/100
=49/100
A=1/2!+1/2!-1/3!+...+1/99!-1/100
=1/2-1/100
=49/100
\(G=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\) CMR G nhỏ hơn \(\frac{3}{4}\)
Cho A= \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
CMR A bé hơn 1.
Làm nhanh giùm mình!!
Cho A=\(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.\frac{99}{100}\)
CMR \(\frac{1}{15}\)bé hơn A bé hơn \(\frac{1}{10}\)
Làm nhanh giùm mình!
chứng tỏ A nhỏ hơn \(\frac{1}{2}\)
A=\(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+........+\frac{1}{100^2}\)
CMR:
a, \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
b, \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+.....+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
CMR
a)\(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
b)\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+....+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
CMR: \(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}+\frac{1}{100^2}<1\frac{3}{4}\)
Cho dãy:\(1\frac{1}{3};1\frac{1}{8};1\frac{1}{15};....\)
\(CMR:\)Tổng của 100 số hạng đầu tiên nhỏ hơn \(\frac{1}{4}\)
CMR:
a, \(100-\left(1+\frac{1}{2}+\frac{1}{3}+..+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+..+\frac{99}{100}\)
b, \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+..+\frac{1}{200}\right)=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
Giải nhanh giùm mình nhé!!!!!!!!!!!!!!