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29 tháng 12 2016

\(B=\frac{3^n+1}{3^n}=1+\frac{1}{3^n}=C+D\) 

B có 98 số hạng => C=98

\(D=\frac{1}{3}+\frac{..1}{3^{97}}+\frac{1}{3^{98}}\) 

3.D=1+1/3+....+1/3^97

tRỪ CHO NHAU

2D=1-1/3^98

\(C=\frac{1}{2}-\frac{1}{2.3^{98}}< \frac{1}{2}\)

\(B=98+\frac{1}{2}-\frac{1}{2.3^{98}}< 99< 100\) có lẽ đề lấy 100 co chẵn. hay cộng nhầm ai tets hộ cái

19 tháng 10 2015

\(B=\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+...+\frac{3^{98}+1}{3^{98}}\)

\(=\frac{3+1}{3}+\frac{3^2+1}{3^2}+\frac{3^3+1}{3^3}+...+\frac{3^{98}+1}{3^{98}}\)

\(=1+\frac{1}{3}+1+\frac{1}{3^2}+1+\frac{1}{3^3}+...+1+\frac{1}{3^{98}}\)

\(=98+\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right)\)

\(\text{Đặt }A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\)

\(\text{rút gon cái A thì dc: }A=\frac{1}{3^{98}}-1\Rightarrow B=98+\frac{1}{3^{98}}-1=97+\frac{1}{3^{98}}\)

\(

12 tháng 7 2019

A=\(\frac{4}{3}+\frac{10}{3^2}+...+\frac{3^{98}+1}{3^{98}}\)

=>  A>\(\frac{3}{3}+\frac{9}{9}+...+\frac{3^{98}}{3^{98}}\)  = 1+1+..+1 =98

A=\(\frac{3}{3}+\frac{9}{9}+...+\frac{3^{98}}{3^{98}}\) +\(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)> 1+1+..+1 = 98 

Đặt  B = \(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)

=> 3B  =  \(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}\)

=>2B =    1-\(\frac{1}{3^{98}}\)              <1    

=> B<1

=>A<99

=>98<A<99            

20 tháng 7 2019

mk doan la` de sai, sua: \(\frac{3^9-2^3.3^7+2^{10}.3^2-2^{13}}{3^{10}-2^2.3^7+2^{10}.3^3-2^{12}}\)

\(=\frac{3^7.\left(3^2-2^3\right)+2^{10}.\left(3^2-2^3\right)}{3^7.\left(3^3-2^2\right)+2^{10}.\left(3^3-2^2\right)}=\frac{3^7+2^{10}}{\left(3^7+2^{10}\right).24}=\frac{1}{24}\)

18 tháng 3 2018

\(B=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{\frac{99}{1}+\frac{98}{2}+\frac{97}{3}+...+\frac{1}{99}}\)

\(B=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{\left(\frac{98}{2}+1\right)+\left(\frac{97}{3}+1\right)+...+\left(\frac{1}{99}+1\right)+1}\)

\(B=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{\frac{100}{2}+\frac{100}{3}+...+\frac{100}{99}+\frac{100}{100}}\) 

\(B=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)}\)

\(B=\frac{1}{100}\)

19 tháng 3 2016

M=100

Xét tử N

92-(1/9)-(2/10)-(3/11)- ... -(90/98)-(91/99)-(92/100)

=(1+1+1+...+1)-(1/9)-(2/10)-(3/11)- ... -(90/98)-(91/99)-(92/100)

=1-(1/9)+1-(2/10)+1-(3/11)+......+1-(90/98)+1-(91/99)+1-(92/100)

=(8/9)+(8/10)+(8/11)+ ...+ (8/98)+(8/99)+(8/100)

=8.[(1/9)+(1/10)+(1/11)+...+(1/98)+(1/99)+(1/100)]

=40[(1/45)+(1/50)+(1/55)+...+(1/495)+(1/500)]

=>N=40

=>M/N=5/2