Đáp án:

 

Giải thích các bước giải:

A= 1/2+1/2^2+...+1/2^100

2A=1+1/2+1/2^2+...+1/2^99

A-2A=1/2^100 -1

-A=-1⇒A=1

B=1+1/5+1/5^2+...+1/5^500

5B=5+1+1/5+1/5^2+...+1/5^499

B-5B=1/5^500 -5=-5

-4B=-5

B=5/4

2. M=$\frac{1}{3}$ *($\frac{1}{20}$- $\frac{1}{23}$ +$\frac{1}{23}$ -$\frac{1}{26}$ +...+$\frac{1}{77}$ -$\frac{1}{80}$ )

    M=$\frac{1}{3}$* ($\frac{1}{20}$- $\frac{1}{80}$ )

    M=$\frac{1}{3}$* $\frac{3}{80}$ 

    M=$\frac{1}{80}$< $\frac{1}{9}$ 

Đáp án:

 

Giải thích các bước giải:

 $A=\dfrac{1}{2}+\dfrac{1}{2^2}}+...+\dfrac{1}{2^{100}}$

$ $

$⇒\dfrac{1}{2}.A=\dfrac{1}{2^{2}}+\dfrac{1}{2^{3}}+...+\dfrac{1}{2^{101}}$

$ $

$⇒A-\dfrac{1}{2}.A=(\dfrac{1}{2}+\dfrac{1}{2^2}}+...+\dfrac{1}{2^{100}})-(\dfrac{1}{2^{2}}+\dfrac{1}{2^{3}}+...+\dfrac{1}{2^{101}})$

$ $

$⇒\dfrac{1}{2}.A=\dfrac{1}{2}-\dfrac{1}{2^{101}}$

$ $

$⇒\dfrac{1}{2}.A=\dfrac{2^{100}-1}{2^{101}}$

$ $

$⇒A=\dfrac{2^{101}-2}{2^{101}}$

$ $

$ $

$ $

$B=1+\dfrac{1}{5}+\dfrac{1}{5^{2}}+\dfrac{1}{5^{3}}+...+\dfrac{1}{5^{500}}$

$ $

$⇒\dfrac{1}{5}.B=\dfrac{1}{5}+\dfrac{1}{5^{2}}+...+\dfrac{1}{5^{501}}$

$ $

$⇒B-\dfrac{1}{5}.B=(1+\dfrac{1}{5}+\dfrac{1}{5^{2}}+\dfrac{1}{5^{3}}+...+\dfrac{1}{5^{500}})-(\dfrac{1}{5}+\dfrac{1}{5^{2}}+...+\dfrac{1}{5^{501}})$

$ $

$⇒\dfrac{4}{5}.B=1-\dfrac{1}{5^{501}}$

$ $

$⇒\dfrac{4}{5}.B=\dfrac{5^{501}-1}{5^{501}}$

$ $

$⇒B=\dfrac{5^{502}-5}{5^{501}.4}$

$ $

$ $

$ $

$ $

$M=\dfrac{1}{20.23}+\dfrac{1}{23.26}+...+\dfrac{1}{77.80}$

$ $

$⇒M=\dfrac{1}{3}.(\dfrac{3}{20.23}+\dfrac{3}{23.26}+...+\dfrac{3}{77.80})$

$ $

$⇒M=\dfrac{1}{3}.(\dfrac{1}{20}-\dfrac{1}{80})$

$ $

$⇒M=\dfrac{1}{3}.\dfrac{3}{80}$

$ $

$⇒M=\dfrac{1}{80}<\dfrac{1}{9}$   (đpcm)