tính(1+2/1)(1+2/2)(1+2/3).....(1+2/26)(1+2/27)
(1-1/2^2)(1-1/3^2)(1-1/4^2)........(1-1/2007^2)
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A =\(\left(1+\frac{2}{1}\right)\left(1+\frac{2}{2}\right)\left(1+\frac{2}{3}\right)\left(1+\frac{2}{4}\right)...\left(1+\frac{2}{26}\right)\left(1+\frac{2}{27}\right)\)
\(=\frac{3}{1}.\frac{4}{2}.\frac{5}{3}.\frac{6}{4}....\frac{28}{26}.\frac{29}{27}=\frac{28.29}{1.2}=14.29=406\)
\(=\dfrac{3}{2}\cdot\dfrac{4}{2}\cdot\dfrac{5}{3}\cdot\dfrac{6}{4}\cdot...\cdot\dfrac{27}{25}\cdot\dfrac{28}{26}\cdot\dfrac{29}{27}\)
\(=\dfrac{1}{4}\cdot28\cdot29=7\cdot29=203\)
\(\frac{M}{N}=\frac{\frac{1}{2007}+\frac{2}{2006}+......+\frac{2006}{2}+\frac{2007}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.......+\frac{1}{2006}+\frac{1}{2007}}\)
\(\frac{M}{N}=\frac{\frac{1}{2007}+1+\frac{2}{2006}+1+.......+\frac{2007}{1}+1+\frac{2008}{2008}-2008}{\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+.....+\frac{1}{2}}\)
\(\frac{M}{N}=\frac{\frac{2008}{2007}+\frac{2008}{2006}+....+\frac{2008}{1}+\frac{2008}{2008}-2008}{\frac{1}{2008}+........+\frac{1}{2}}\)
đến đây là ra rùi ha
mk chỉ pik câu C thui
C= (1 - 1/2).(1 - 1/3).(1 - 1/4).......(1 - 4/4)....(1-1/2007)
C= (1 - 1/2).(1 - 1/3)..........0........(1 - 1/2007)
C= 0
mình chỉ biết câu b thôi:
Ta biến đổi vế phải :
1-1/2+1/3-1/4+.....+1/49-1/50
=(1+1/3+1/5+....+1/49)-(1/2+1/4+1/6+.......+1/50)
=(1+1/2+1/3+.....+1/49+1/50)-2(1/2+1/4+1/6+......+1/50)
=(1+1/2+...+1/50)-(1+1/2+1/3+....+1/25)
=1/26+1/27+.......+1/50
Vậy 1/26+1/27+1/28+.....+1/50=1-1/2+1/3-1/4+......+1/49-1/50