Các câu hỏi tương tự
CMR 1+1/2+1/3+.......+1/2^1999 >1000
Câu hỏi: CMR: 1+1/2+1/3+....1/2^1999 >1000
C ={(1+(1999/1))(1+(1999/2))(1+(1999/3))+...+(1+(1999/1000))}/{(1+(1000/1))(1+(1000/2))(1+(1000/3))...(1+(1000/1999))}
cho M = 1 + 1/2 + 1/3 +...+1/2 mũ 1999
Chứng minh M > 1000
giúp mk với nha gấp lắm á
cảm ơn mn
1.Tính C=\(\frac{\left(1+\frac{1999}{1}\right)\left(1+\frac{1999}{2}\right)\left(1+\frac{1999}{3}\right)...\left(1+\frac{1999}{1000}\right)}{\left(1+\frac{1000}{1}\right)\left(1+\frac{1000}{2}\right)\left(1+\frac{1000}{3}\right)...\left(1+\frac{1000}{1999}\right)}\)
1)CMR :1+12+1/3+..+1/2^1999<1000
2)tím số tự nhiên x biết :1/3+1/3+1/10+...+1/x(x+1)=2013/2015
\(A=\frac{\left(1+\frac{1999}{1}\right)\left(1+\frac{1999}{2}\right)\left(1+\frac{1999}{3}\right)...\left(1+\frac{1999}{1000}\right)}{\left(1+\frac{1000}{1}\right)\left(1+\frac{1000}{2}\right)\left(1+\frac{1000}{3}\right)...\left(1+\frac{1000}{1999}\right)}\)
hỏi a = ?
A = (1 + 1999/1)(1 + 1999/2)......(1 + 1999/1000)
B = ( 1 + 1000/1)(1 + 1000/2)......(1 + 1000/1999)
Tính A/B
<5x-6>*<1999 mũ 2+2*1999+1>=<4*10 mũ 3>mũ 2
giúp mình với