Ta có: \(1+\frac{1}{2}+\frac{1}{3}+...\frac{1}{2^{1999}}=1+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{2^2}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{2^3}\right)+...\left(\frac{1}{2^{1998+1}}+...\frac{1}{2^{1999}}\right)>1+\frac{1}{2}+\frac{1}{2^2.2}+\frac{1}{2^3.2^2}+...+\frac{1}{2^{1999}-2^{1998}}=1+\frac{1}{2}.1999=1000,5>1000\)
1.Tính C=\(\frac{\left(1+\frac{1999}{1}\right)\left(1+\frac{1999}{2}\right)\left(1+\frac{1999}{3}\right)...\left(1+\frac{1999}{1000}\right)}{\left(1+\frac{1000}{1}\right)\left(1+\frac{1000}{2}\right)\left(1+\frac{1000}{3}\right)...\left(1+\frac{1000}{1999}\right)}\)
\(A=\frac{\left(1+\frac{1999}{1}\right)\left(1+\frac{1999}{2}\right)\left(1+\frac{1999}{3}\right)...\left(1+\frac{1999}{1000}\right)}{\left(1+\frac{1000}{1}\right)\left(1+\frac{1000}{2}\right)\left(1+\frac{1000}{3}\right)...\left(1+\frac{1000}{1999}\right)}\)
hỏi a = ?
Tính :
C= \(\frac{\left(1+\frac{1999}{1}\right)\left(1+\frac{1999}{2}\right)...\left(1+\frac{1999}{1000}\right)}{\left(1+\frac{1000}{2}\right)\left(1+\frac{1000}{2}\right)...\left(1+\frac{1000}{1999}\right)}\)
Chứng tỏ :
\(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^{1999}}>1000\)
Chứng minh rằng :
\(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^{1999}}>1000\)
Chứng minh 1 + \(\frac{1}{2}\) + \(\frac{1}{3}\) + ... + \(\frac{1}{2^{1999}}\) > 1000
\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{20}}{\frac{1999}{1}+\frac{1998}{2}+\frac{1997}{3}+...+\frac{1}{1999}}\)
\(E=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}{\frac{1999}{1}+\frac{1998}{2}+\frac{1997}{3}+..+\frac{1}{1999}}\)
Tính \(E=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}{\frac{1999}{1}+\frac{1998}{2}+\frac{1997}{3}+...+\frac{1}{1999}}\)