\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}+\frac{1}{32}-\frac{1}{64}+\frac{1}{64}-\frac{1}{128}\)
\(=1-\frac{1}{128}\)
\(\frac{127}{128}\)
Đặt dãy trên là A
\(\Leftrightarrow2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{64}\)
\(\Leftrightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{128}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{128}\right)\)
\(\Leftrightarrow A=1-\frac{1}{128}+0+0+...+0\)
\(\Leftrightarrow A=\frac{127}{128}\)
thiếu dấu = ở dòng cuối cùng nhé :vv
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}+\frac{1}{32}-\frac{1}{64}+\frac{1}{64}-\frac{1}{128}\)
\(=1-\frac{1}{128}\)
\(=\frac{127}{128}\)
_Chúc bạn học tốt_
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
=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
=\(\frac{127}{128}\)
\(A=\frac{1}{2}+\frac{1}{4}+...+\frac{1}{128}\)
\(\Rightarrow2A=\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{128}\right)\times2\)
\(2A=1+\frac{1}{2}+...+\frac{1}{64}\)
\(\Rightarrow A=2A-A=\left(1+\frac{1}{2}+...+\frac{1}{64}\right)-\)\(\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{128}\right)\)
\(A=1-\frac{1}{128}\)
\(A=\frac{127}{128}\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^7}\)
\(\Rightarrow2A=\frac{1}{2}+\frac{1}{2^2}+..+\frac{1}{2^6}\)
\(\Rightarrow2A-A=\left(\frac{1}{2}+\frac{1}{2^2}+..+\frac{1}{2^6}\right)-\left(\frac{1}{2^2}+\frac{1}{2^3}+..+\frac{1}{2^7}\right)\)
\(\Rightarrow A=\frac{1}{2}-\frac{1}{2^7}\)
