Tuyển Cộng tác viên Hoc24 nhiệm kì 26 tại đây: https://forms.gle/dK3zGK3LHFrgvTkJ6
+ \(5N=1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{98}}\)
\(N=\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{98}}+\dfrac{1}{5^{99}}\)
\(\Rightarrow4N=5N-N=1-\dfrac{1}{5^{99}}\)
\(\Rightarrow N=\dfrac{1}{4}-\dfrac{1}{4\cdot5^{99}}< \dfrac{1}{4}\) ( đpcm )
Tính giá trị biểu thức A , biết rằng A = M : N
Mà M = \(\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+\dfrac{4}{96}+...+\dfrac{97}{3}+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
N = \(\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)
Cho \(S_1=1+\dfrac{1}{5}\)
\(S_2=1+\dfrac{1}{5}+\dfrac{1}{5^2}\)
................................
\(S_n=1+\dfrac{1}{5}+...+\dfrac{1}{5^n}\)
Chứng minh rằng: \(\dfrac{1}{5S_1^2}+\dfrac{1}{5^2S_2^2}+...+\dfrac{1}{5^nS_n^2}< \dfrac{1}{4}\)
Cho A=\(\dfrac{2}{1}.\dfrac{4}{3}.\dfrac{6}{5}.\dfrac{8}{7}.\dfrac{10}{9}...\dfrac{100}{99}\). Chứng minh rằng 12<A<13
Rút gọn ;
D = \(\dfrac{1}{5}-\dfrac{1}{5^2}+\dfrac{1}{5^3}-\dfrac{1}{5^4}+\dfrac{1}{5^5}-\dfrac{1}{5^6}+...+\dfrac{1}{5^{99}}-\dfrac{1}{5^{100}}+\dfrac{1}{6.5^{100}}\)
Cho S=\(\dfrac{1}{5^2}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+\dfrac{4}{5^4}+...+\dfrac{2017}{5^{2017}}+\dfrac{2018}{5^{2018}}\).Chứng minh S<\(\dfrac{1}{3}\)
1.Chứng minh rằng :\(\dfrac{5}{8}< \dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{200}< \dfrac{3}{4}\)
Chứng minh rằng : \(\dfrac{1}{6}< \dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+....+\dfrac{1}{100^2}< \dfrac{1}{4}\)
Chứng minh rằng : \(\dfrac{1}{6}< \dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}\)
1. Chứng minh rằng:
\(\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{99}{100}< 1\)
2. Chứng minh rằng:
\(\dfrac{1.2-1}{2!}+\dfrac{2.3-1}{3!}+\dfrac{3.4-1}{4!}+...+\dfrac{99.100-1}{100!}< 2\)
