Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
F=đã cho
=>1/2F=1/4+1/8+1/16+...+1/8192
=>F-1/2F=1/2-1/8192
=>1/2F=1/2-1/8192
=>F=1-1/4096
=>F=4095/4096
Vậy......
Ta có : \(F=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+......+\frac{1}{4096}\)
\(\Rightarrow2F=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.....+\frac{1}{2048}\)
\(\Rightarrow2F-F=1-\frac{1}{4096}\)
\(\Rightarrow F=\frac{4095}{4096}\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}+\frac{1}{32}\)
\(=1-\frac{1}{32}\)
\(=\frac{31}{32}\)
k mk nha
\(=\left(\frac{1}{2}+\begin{cases}1\\4\end{cases}+\frac{1}{8}\right)+\left(\frac{1}{16}+\frac{1}{32}\right)\\ =\frac{7}{8}+\frac{3}{32}\\ =\frac{31}{32}\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}\)
\(=1-\frac{1}{32}\)
\(=\frac{31}{32}\)
có tử = 1 thì k bt à nha
Đặt: \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(\Rightarrow\)\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)
\(\Rightarrow\)\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)\)
\(\Rightarrow\)\(A=1-\frac{1}{32}=\frac{31}{32}\)
mà \(A=\frac{1}{x}\)
nên \(\frac{1}{x}=\frac{31}{32}\)
\(\Rightarrow\)\(x=\frac{32}{31}\)
Vậy...
Cách 1:
B=1/2+1/4+1/8+1/16+1/32+1/64
B=1-1/2 + 1/2-1/4 + 1/4-1/8 +1/8-1/16 + 1/16-1/32 + 1/32-1/64
B=1-1/64
B=63/64
Cách 2:
B=1/2+1/4+1/8+1/16+1/32+1/64
B=1/21+1/22+1/23+1/24+1/25+1/26
2B=1+1/21+1/2^2+1/2^3+1/2^4+1/2^5
2B-B=1-1/2^6
B=1-1/64
B=63/64
Đặt A = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64
2A = 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32
2A - A = (1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32) - (1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64)
A = 1 - 1/64
A = 63/64
= 1 - 1/2+ 1/2- 1/4 +1/4 - 1/8 +1/8 -1/16 +1/16 -1/32 +1/32 -1/64 +1/64 - 1/128
= 1-1/128
=127/128
\(\frac{1}{2}\)+ \(\frac{1}{4}\)+ \(\frac{1}{8}\)+ \(\frac{1}{16}\)+ \(\frac{1}{32}\)+ \(\frac{1}{64}\)+ \(\frac{1}{128}\)= \(\frac{64}{128}\)+ \(\frac{32}{128}\)+ \(\frac{16}{128}\)+ \(\frac{8}{128}\)+ \(\frac{4}{128}\)+ \(\frac{2}{128}\)+ \(\frac{1}{128}\).
= \(\frac{127}{128}\).
\(\frac{1}{2}\)+ \(\frac{1}{4}\)+ \(\frac{1}{8}\)+ \(\frac{1}{16}\)+ \(\frac{1}{32}\)+ \(\frac{1}{64}\)+ \(\frac{1}{128}\)
= \(1\)- \(\frac{1}{2}\)+ \(\frac{1}{2}\)- \(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{8}\)+ \(\frac{1}{8}\)- \(\frac{1}{16}\)+ \(\frac{1}{16}\)- \(\frac{1}{32}\)+ \(\frac{1}{32}\)- \(\frac{1}{64}\)+ \(\frac{1}{64}\)- \(\frac{1}{128}\)
= \(1\)- \(\frac{1}{128}\)
= \(\frac{127}{128}\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{128}-\frac{1}{256}\)
=\(1-\frac{1}{256}\)
=\(\frac{255}{256}\)
1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256
= 128/256 + 64/256 + 32/256 + 16/256 + 8/256 + 4/256 + 2/128 + 1/256
= 255/256
\(y=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(\frac{1}{2}y=\frac{1}{2}\times\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)\)
\(\frac{1}{2}y=\frac{1}{2}\times1+\frac{1}{2}\times\frac{1}{2}+...+\frac{1}{2}\times\frac{1}{32}\)
\(\frac{1}{2}y=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(y-\frac{1}{2}y=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{32}\right)-\left(\frac{1}{2}+\frac{1}{4}+..+\frac{1}{64}\right)\)
\(\left(1-\frac{1}{2}\right)y=1-\frac{1}{64}\)
\(\frac{1}{2}y=\frac{63}{64}\)
\(y=\frac{63}{64}\div\frac{1}{2}=\frac{63}{32}\)