Các câu hỏi tương tự
chúng minh rằng
1) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}<1\)
2) \(\frac{49}{100}<\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}<1\)
CMR
\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}<2\)
a, A=\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}< 1\)
b, B=\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{99.100}< 2\)
c, C=\(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{63}< 6\)
d, D=\(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}....\frac{9999}{10000}< \frac{1}{100}\)
Bài 9: tính
a, A= 1+2+3+4+....+100
b,B=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+........+\frac{1}{99.100}\)
c, C=\(\frac{10}{56}+\frac{10}{140}+\frac{10}{200}+.......+\frac{1}{1400}\)
\(CMR:\) \(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\) \(< 2\)
\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^{^2}}{3.4}...\frac{99^2}{99.100}.\frac{100^2}{100.101}\)
a) Rút gọn: \(\frac{9.25-63}{3.30+153}\)
b) Chứng tỏ rằng:
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}< \frac{1}{2}\)
So sánh :
\(A=\frac{1}{1.2^2}+\frac{1}{2.3^2}+\frac{1}{3.4^2}+...+\frac{1}{98.99^2}+\frac{1}{99.100^2}\) và \(B=\frac{5}{12}\)
Bài 15.
a) So sánh \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)và 1
b) Cho biểu thức A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{1000}}.\)Chứng tỏ A < 1