Ta có:\(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{200}>\dfrac{1}{200}+\dfrac{1}{200}+...+\dfrac{1}{200}=\dfrac{100}{200}=\dfrac{1}{2}\)
Lại có:
\(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{200}< \dfrac{1}{101}+\dfrac{1}{101}+...+\dfrac{1}{101}=\dfrac{100}{101}\)
Vậy ...
Những dãy trên đều có 100 số hạng.
Cho \(A=\dfrac{10^{101}-1}{10^{102}-1}\) và \(B=\dfrac{10^{100}+1}{10^{101}+1}\)
So sánh A và B
Cho A=1+\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2^{100}-1}\)
Chứng minh rằng 50<A<100
cho \(\dfrac{1}{p}=\dfrac{1}{2}\left(\dfrac{1}{m}+\dfrac{1}{n}\right)\)(với m,n,p#0,n#p)Chứng minh rằng\(\dfrac{m}{n}=\dfrac{m-p}{p-m}\)
1) Tính
\(\dfrac{7^4.3-7^3}{7^4.6-7^3.2}\) ; \(\dfrac{10^3+5.10^2+5}{6^3+3.6^2+3^2}\) ; \(E=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\)
2) Tìm x biết
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{x.\left(x+1\right)}\) ; \(3^{x+1}+3^{x+3}=810\)
MN ƠI ! GIÚP MIK VS > . <
Cho \(A=1+\dfrac{1}{4}+\dfrac{1}{9}+...+\dfrac{1}{n^2}\) với n là số tự nhiên. Chứng minh rằng \(A< \dfrac{7}{4}\).
Cho \(A=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{98}+\left(\dfrac{1}{2}\right)^{99}\)
Chứng minh : A<1
Tính A= \(\dfrac{1}{2}+\dfrac{1}{2}.\left(1+2\right)+\dfrac{1}{3}.\left(1+2+3\right)+...+\dfrac{1}{100}.\left(1+2+3+...+100\right)\)
Cho A= \(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+.......+\dfrac{1}{3^{2018}}+\dfrac{1}{3^{2019}}\) . Chứng minh rằng: A<\(\dfrac{1}{2}\)
AI LÀM ĐƯỢC MÌNH SẼ TICK!!!
Cho \(A=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{98}+\left(\dfrac{1}{2}\right)^{99}\)
Chứng minh : \(A< 1\)
