\(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}=\frac{32+16+8+4+2+1}{128}=\frac{63}{128}\)
Đặt A=1/4+1/8+1/16+1/32+1/64+1/128.
\(A=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\)
\(2A=\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\right)\)
\(2A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\)
\(2A-A=\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\right)-\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\right)\)
\(A=\frac{1}{2}-\frac{1}{2^7}\)
A=1/4+1/8+1/16+1/32+1/64+1/128
Ax2-A=(1/4+1/8+1/16+1/32+1/64+1/128)x2-(1/4+1/8+1/16+1/32+1/64+1/128)
Ax2-A=(1/2+1/4+1/8+1/16+1/32+1/64)-(1/4+1/8+1/16+1/32+1/64+1/128)
Ax2-A=1/2-1/4+1/4-1/8+1/8-1/16+1/16-1/32+1/32-1/64+1/64-1/128
A=1/2-1/128
A= 63/128.
1/4+1/8+1/16+1/32+1/64+1/128 = 63,128
Mong mọi ng k ạ !!! Ai đi qua xin để lại 1 k !!!!
ஜ۩۞۩ஜ Ủng hộ tui nha ஜ۩۞۩ஜ