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Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)\)
\(A=1-\frac{1}{32}=\frac{31}{32}\)
\(\frac{1}{2}\)+ \(\frac{1}{4}\) + \(\frac{1}{8}\) + \(\frac{1}{16}\) + \(\frac{1}{32}\)
= [ 1 - \(\frac{1}{2}\)] + [ \(\frac{1}{2}\) - \(\frac{1}{4}\)] + [ \(\frac{1}{4}\) - \(\frac{1}{8}\)] + [ \(\frac{1}{8}\) - \(\frac{1}{16}\)] + [ \(\frac{1}{16}\) - \(\frac{1}{32}\)]
Xóa bỏ các phân số trùng lặp , ta được tổng của dãy số là :
1 - \(\frac{1}{32}\) = \(\frac{31}{32}\)
Đ/S :\(\frac{31}{32}\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}=\frac{32}{64}+\frac{16}{64}+\frac{8}{64}+\frac{4}{64}+\frac{2}{32}+\frac{1}{64}\)
\(\frac{32+16+8+4+2}{64}=\frac{62}{64}=\frac{31}{32}\)
Tk mh nhé , mơn nhìu !!!
~ HOK TỐT ~
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)\(+\frac{1}{64}\)
= 32/64 + 16/64 + 8/64 + 4/64 + 2/64 + 1/64
= 63/64
Đặt A = \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
2A = \(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
2A - A = \(1-\frac{1}{64}\)
=> A = \(\frac{63}{64}\)
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 : \(F=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+......+\frac{1}{4096}\)
\(\Rightarrow2F=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.....+\frac{1}{2048}\)
\(\Rightarrow2F-F=1-\frac{1}{4096}\)
\(\Rightarrow F=\frac{4095}{4096}\)
Ta có: \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}\)
=>2A=\(1+\frac{1}{2^2}+...+\frac{1}{2^4}+\frac{1}{2^5}\)
=>2A-A=(\(1+\frac{1}{2^2}+...+\frac{1}{2^4}+\frac{1}{2^5}\))--(\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}\))
=>A=\(1-\frac{1}{2^6}\)
=>A=\(\frac{63}{64}\)
gọi biểu thức là A
A=1/2+1/4+1/8+...+1/2048=1/2+1/2^2+1/2^3+...+1/2^10
=>2A=1+1/2+1/2^2+...+1/2^9
=>A=2A-A(bạn đặt cột dọc ra rồi sẽ thấy:1/2-1/2=0;1/2^2-1/2^2=0;...)Ta được kết quả bằng 1+1/2^10
Đặt A =1/2 + 1/4 + 1/8 + ...+ 1/1024 + 1/2048
A= 1/2 + 1/2^2 + 1/2^3+...+ 1/2^10 + 1/2^11
2A= 1 +1/2 + 1/2^2 +...+ 1/2^9 + 1/2^10
2A-A= (1 +1/2 + 1/2^2 +...+ 1/2^9 + 1/2^10) - (1/2 + 1/2^2 + 1/2^3+...+ 1/2^10 + 1/2^11)
A= 1+1/2 + 1/2^2 +...+ 1/2^9 + 1/2^10 - 1/2 - 1/2^2 - 1/2^3 - ...- 1/2^10 - 1/2^11
A= 1- 1/2^11
A= 2047/ 2048
=\(\frac{31}{32}\)
Đặt biểu thức trên là A
Ta có:
\(2A-A=A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)=1-\frac{1}{32}=\frac{31}{32}\)