Tham khảo tại link sau : olm.vn/hoi-dap/question/687403.html
Tham khảo tại link sau : olm.vn/hoi-dap/question/687403.html
CMR:
\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{2019\times2020}< 1\)
\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< \frac{3}{4}\)
\(C=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}< 2\)
Cmr:\(50< 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2^{100-1}}< 100\)
CMR: \(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{50^2}< \frac{173}{100}\)
CMR:
a,\(100\left(1+\frac{1}{2}+\frac{1}{3}+..........+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+........+\frac{99}{100}\)
CMR :
a , A = \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+.....+\frac{19}{9^2.10^2}< 1\)
b , B = \(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+......+\frac{100}{3^{100}}< \frac{3}{4}\)
c, C = \(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right).....\left(\frac{1}{100^2}-1\right)< \frac{1}{2}\)
CMR
a)A=\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+....+\frac{100}{3^{100}}< \frac{3}{4}\)
b)B=\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+.....+\frac{100}{4^{100}}< \frac{4}{9}\)
Cho \(P=\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+..+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..+\frac{1}{100}}\)và \(Q=\frac{92-\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-..-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+..+\frac{1}{500}}\)
a)Tính P,Q b) Tính tỉ số % của P và 3Q
a)A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}< 1\)
b)B=\(\frac{1}{3}+\left(\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^3+...+\left(\frac{1}{3}\right)^{100}< \frac{1}{2}\)
c)\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\)
d)A=\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}.CMR\frac{7}{12}< A< \frac{5}{6}\)
AI ĐÚNG MINK \(\left(TICK\right)\)CHO (làm đc trên 2 câu)
Cho A=\(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}\)
CMR \(\frac{1}{6}< A< \frac{1}{4}\)