Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)\)
\(A=1-\frac{1}{32}=\frac{31}{32}\)
1/2+1/4+1/8+1/16+1/32
ta có:
1/2=1-1/2
1/4=1/2-1/4
1/8=1/4-1/8
.........
Cứ thế,ta có dãy tính:
1-1/2+1/2-1/4+1/4-1/8+1/8-1/16+1/16-1/32
=1-1/32
31/32
Đặt biểu thức trên là A ta có
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\)\(\frac{1}{32}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)
\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)\)
\(\Rightarrow A=1-\frac{1}{32}\)
\(A=\frac{31}{32}\)
C1:\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(=\frac{16}{32}+\frac{8}{32}+\frac{4}{32}+\frac{2}{32}+\frac{1}{32}\)
\(=\frac{16+8+4+2+1}{32}\)
\(=\frac{31}{32}\)
C2: gọi \(S=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(\Leftrightarrow2S=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)
\(\Leftrightarrow2S-S=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)\)
\(\Leftrightarrow S=1-\frac{1}{32}=\frac{31}{32}\)
\(S=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(2S=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)
\(2S-S=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)\)
\(S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}\)
\(S=1-\frac{1}{32}\)
\(S=\frac{31}{32}\)
Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^5}\)
\(\Rightarrow2A=1+\frac{1}{2}+...+\frac{1}{2^4}\)
\(\Rightarrow2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^4}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^5}\right)\)
\(\Rightarrow A=1-\frac{1}{2^5}\)
Đặt \(B=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(2B=\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)\times2\)
\(2B=\frac{1}{2}\times2+\frac{1}{4}\times2+\frac{1}{8}\times2+\frac{1}{16}\times2+\frac{1}{32}\times2\)
\(2B=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)
\(2B-B=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)\)
\(B=1-\frac{1}{32}\)
\(B=\frac{31}{32}\)
