Tuyển Cộng tác viên Hoc24 nhiệm kì 26 tại đây: https://forms.gle/dK3zGK3LHFrgvTkJ6
\(A=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+........+\frac{100}{2^{100}}\)
\(\Rightarrow2A=1+\frac{2}{2}+\frac{3}{2^2}+..........+\frac{100}{2^{99}}\)
\(\Rightarrow2A-A=A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+......+\frac{1}{2^{99}}+\frac{100}{2^{100}}\)
\(\Rightarrow2A=2+1+\frac{1}{2}+\frac{1}{2^2}+......+\frac{1}{2^{98}}+\frac{100}{2^{99}}\)
\(\Rightarrow2A-A=A=2-\frac{100}{2^{99}}< 2\)
Vậy \(A< 2\)
Cho A=1+\(\frac{3}{2^3}+\frac{4}{2^4}+\frac{5}{2^5}+...+\frac{100}{2^{100}}\)
CMR:A không phải là số nguyên
\(A=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+.......+\frac{99}{100!}.CMR:A
chứng tỏ rằng
C = \(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+...+\frac{1}{2^{99}}-\frac{1}{2^{100}}< \frac{1}{3}\)
D = \(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}< \frac{3}{4}\)
Chứng minh rằng:
a. \(\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+\frac{4}{3^5}+...+\frac{99}{3^{100}}+\frac{100}{3^{101}}< \frac{1}{4}\)
b.\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
c.\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{1}{16}\)
d. \(\frac{1}{5^2}-\frac{2}{5^3}+\frac{3}{5^4}-\frac{4}{5^5}+...+\frac{99}{5^{100}}-\frac{100}{5^{101}}< \frac{1}{36}\)
Chứng minh rằng:
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\) \(B=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}+\frac{101}{3^{101}}< \frac{3}{4}\)
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=\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+.....+\frac{100}{4^{100}}< \frac{4}{9}\)
A=\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)\(< \frac{3}{4}\)
B=\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}< 1\)
C=\(1+3+3^2+3^3+...+3^{100}\)
D=\(2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
Gửi
TNs tao cuồng:c/m \(B=\frac{1}{2}+\frac{1}{2^2}+\frac{3}{2^3}+....+\frac{100}{2^{100}}<2\)Ta có:\(2B=1+\frac{1}{2}+\frac{3}{2^2}+....+\frac{100}{2^{99}}\)\(\Rightarrow2B-B=B=1+\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{99}}\right)-\frac{100}{2^{100}}\)(*)c/m \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}<1\)Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{99}}\)\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{99}}\right)\)\(\Rightarrow A=1-\frac{1}{2^{99}}<1\)do đó \(B=1+A-\frac{100}{2^{100}}\Rightarrow B<2-\frac{100}{2^{100}}<2\left(đpcm\right)\)