Đặt A=1/2−1/4+1/8−1/16+1/32−1/64A=1/2−1/4+1/8−1/16+1/32−1/64
2A=1−1/2+1/4−1/8+1/16−1/322A=1−1/2+1/4−1/8+1/16−1/32
3A=2A+A=1−1/64<1⇒A<1/3

mồm ngày trc thì dùng nick này bảo mình là kaitoru xong nói qua lại bt hết điêu ta bốc phét

đặt A làm tên biểu thức trên . Ta có :

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

\(2A=2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)\)

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

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)\)

\(A=1-\frac{1}{64}\)

\(\Rightarrow A=\frac{63}{64}\)

= (1/2 + 1/64) x (6 : 2) = 99/64
 

      A  =          \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) +    \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\)+.....+  \(\dfrac{1}{134}\)+ \(\dfrac{1}{268}\)

A \(\times\) 2 =  1  +  \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\)   + \(\dfrac{1}{8}\)  + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\) +.....+ \(\dfrac{1}{134}\)

A \(\times\) 2 - A = 1 - \(\dfrac{1}{268}\)

A \(\times\) ( 2  - 1) = \(\dfrac{267}{268}\)

A = \(\dfrac{267}{268}\)

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
 

\(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+...+\frac{1}{512}+\frac{1}{1024}=??????????\)

\(< =>1+\frac{1}{1\cdot2}+\frac{1}{2\cdot2}+\frac{1}{2\cdot4}+...+\frac{1}{2\cdot256}+\frac{1}{2\cdot512}\)

\(< =>1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{2}-\frac{1}{256}+\frac{1}{2}-\frac{1}{512}\)

\(< =>1+\frac{1}{1}-\frac{1}{512}\)

\(< =>\frac{1023}{512}\)

chuc ban hoc tot nhe :))

b: A=1/3+1/9+...+1/3^10

=>3A=1+1/3+...+1/3^9

=>A*2=1-1/3^10=(3^10-1)/3^10

=>A=(3^10-1)/(2*3^10)

c: C=3/2+3/8+3/32+3/128+3/512

=>4C=6+3/2+...+3/128

=>3C=6-3/512

=>C=1023/512

d: A=1/2+...+1/256

=>2A=1+1/2+...+1/128

=>A=1-1/256=255/256

\(\frac{1}{2}\)+ \(\frac{1}{4}\)+ \(\frac{1}{16}\)+ \(\frac{1}{32}\)+ \(\frac{1}{64}\)+ \(\frac{1}{128}\)= \(\frac{123}{234}\)

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

S=1-1/2+1/2-1/4+1/4-1/8+1/8-1/16+1/16-1/32+1/32-1/64

S=1-1/64

S=63/64

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

2S = 1/2 x 2 +1/4 x 2 + 1/8 x 2 + 1/16 x 2 + 1/32 x 2 + 1/64 x 2

2S =1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32

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

S = 1 - 1/64

S = 63/64