Lời giải:

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

$\Rightarrow 2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}$

$\Rightarrow A=2A-A=1-\frac{1}{32}< 1-\frac{1}{2004}$

Hay $A< \frac{2003}{2004}$

Hay $A< B$

Em xem lại đề xem đã đúng chưa?

em thưa cô đề này là em dc cô giáo học thêm cho và cô chưa dậy nên em nghĩ là đúng đề rồi ạ

S=1-3+3²-...+3⁹⁸-3⁹⁹ (1)

=>3S=3-3²+3³+...+3⁹⁹-3¹⁰⁰(2)

Từ 1 và 2 =>4S=1-3¹⁰⁰

Do S chia hết cho -20 =>4S chia hết cho -20=>4S chia hết cho 4=>1-3¹⁰⁰ chia hết cho 4

=>3¹⁰⁰ chia 4 dư 1

tick đúng mình nha rùi mình giải tiếp dài quá

sao?

1/2 + 1/4 + 1/8 + 1/16+ 1/32 + 1/64 + 1/128 

= 64/ 128 + 32/128 + 16/128 +8/128 + 4/128 +2/128 + 1/128

= ( 64 + 32 + 16 + 8 + 4 + 2 + 1 ) /128

= 127/ 128

= 1 - 1/2 + 1/2 - 1/4 + 1/4 - ............ + 1/64 - 1/128

= 1 - 1/128 

= 127/128

k nha bn

Đặt A= 1/2+1/4+1/8+1/16+1/32+1/64+1/128

     2A=1+1/2+1/4+1/8+1/16+1/32+1/64

     2A-A=(1+1/2+1/4+1/8+1/16+1/32+1/64)-(1/2+1/4+1/8+1/16+1/32+1/64+1/128)

     A=1-1/128=127/128
 

(1/2+1/8)+(1/4+1/16)+(1/8+1/32)+1/64

=1/10+1/20+1/40+1/64

=61/320

lấy 1/64 làm mẫu xố chung

1/64+2/64+4/64+8/64+12/64+32/64=59/64

MTC: 64

=> \(\frac{32}{64}+\frac{16}{64}+\frac{8}{64}+\frac{4}{64}+\frac{2}{64}+\frac{1}{64}\)

= \(\frac{63}{64}\)

         A =    \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\)+ \(\dfrac{1}{32}\)+\(\dfrac{1}{64}\)+\(\dfrac{1}{128}\)

A\(\times\) 2 =  1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\)+ \(\dfrac{1}{32}\)+ \(\dfrac{1}{64}\) 

A \(\times\) 2 - A = 1 - \(\dfrac{1}{128}\)

A\(\times\)(2-1) = \(\dfrac{128-1}{128}\)

A           = \(\dfrac{127}{128}\)

Gọi \(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\) là B

\(B=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\)

\(2\cdot B=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{12}+\dfrac{1}{32}+\dfrac{1}{64}\)

\(2\cdot B-B=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{12}+\dfrac{1}{32}+\dfrac{1}{64}-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\right)\)

\(B=1+\left(\dfrac{1}{2}-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+.....+\dfrac{1}{64}-\dfrac{1}{64}\right)-\dfrac{1}{128}\)

\(B=1+0-\dfrac{1}{128}\)

\(B=1-\dfrac{1}{128}\)

\(B=\dfrac{128}{128}-\dfrac{1}{128}\)

\(B=\dfrac{127}{128}\)