Question

Cho:

A=\(\frac{1}{2}\)+\(\frac{1}{4}\)+\(\frac{1}{8}\)+\(\frac{1}{16}\)+\(\frac{1}{32}\)

B=\(\frac{2003}{2004}\)

So sánh A và B

Answer 1

=> A = \(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}\)

Nhân 2 vế vào A , ta được :

=> 2A = \(1+\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}\) ( 1 )

Trừ ( 1 ) cho A , ta được :

2A - A = \(1+\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^1}-\frac{1}{2^2}-\frac{1}{2^3}-\frac{1}{2^4}-\frac{1}{2^5}\) 

=> A = \(1-\frac{1}{2^5}=1-\frac{1}{32}=\frac{31}{32}\)

Kiểm tra : 2003 : 2004 \(\approx\) 0,999

                31 : 32 \(\approx\)0,968

Vì 0,999 > 0,968 => \(\frac{2003}{2004}>\frac{31}{32}\)

 

Answer 2

A=1/2+1/4+1/8+1/16+1/32

A=16/32+8/32+4/32+2/34+1/34

A=31/34

B>A