Câu 1 : Tính
A= 1/2 + 1/3 + 1/4 +...+1/300
B= 2999/1 + 2998/2 + 2997/3 +...+1/2999
Tính \(\frac{A}{B}\)
Câu 2
C= (1+2012/1)(1+2012/2)....(1+2012/1000)
D=(1+1000/1)(1+1000/2)(1+1000/3)...(1+1000/2012)
Tính \(\frac{C}{D}\)
Câu 3
Cho E=1/1.2 + 1/3.4 + 1/5.6 +...+1/2013/2014
F=1/1008/2014 + 1009/2013 +.....+1/2014.1008
Tính \(\frac{E}{F}\)
Câu 1:
B = \(\frac{2999}{1}+\frac{2998}{2}+\frac{2997}{3}+...+\frac{1}{2999}\)
= \(\frac{3000-1}{1}+\frac{3000-2}{2}+\frac{3000-3}{3}+...+\frac{3000-2999}{2999}\)
= \(\left(\frac{3000}{1}+\frac{3000}{2}+\frac{3000}{3}+...+\frac{3000}{2999}\right)-\left(\frac{1}{1}+\frac{2}{2}+\frac{3}{3}+...+\frac{2999}{2999}\right)\)
= \(3000+3000.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2999}\right)-2999\)
= \(3000\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2999}\right)+\frac{3000}{3000}\)
= \(3000\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{3000}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{3000}}{3000\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{3000}\right)}=\frac{1}{3000}\)
C=2013/1*2014/2*2015/3*...*3012/1000
C=2013*2014*2015*...*3012/1*2*3*...*1000
D=1001/1*1002/2*1003/3*...*3012/2012
D=1001*1002*...*3012/1*2*...*2012
Suy ra C/D=2013*2014*2015*...3012*1*2*...*2012/1*2*3*...*1000*1001*1002*...*3012
( Nhân đảo ngược)
Vậy C/D=1
