\(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(=\frac{4+2+1}{16}+\frac{4+2+1}{128}\)
\(=\frac{7}{16}+\frac{7}{128}\)
\(=\frac{56+7}{128}\)
\(=\frac{63}{128}\)
Nhớ k cho mình với nhé!
A = 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128
A x 2 = 1/4 - ( 1/4 + 1/8 + 1/16 + .............. + 1/64 + 1/128 ) - 1/128
A x 2 = 1/4 - A - 1/128
A x 2 - A = 1/4 - 1/128
A = 1/4 - 1/128
A = 31/128
A = 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128
A x 2 = 1/4 - ( 1/4 + 1/8 + 1/16 + .............. + 1/64 + 1/128 ) - 1/128
A x 2 = 1/4 - A - 1/128
A x 2 - A = 1/4 - 1/128
A = 1/4 - 1/128
A = 31/128
\(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}=\)
\(\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+\left(\frac{1}{8}-\frac{1}{16}\right)+...+\left(\frac{1}{64}-\frac{1}{128}\right)\)
TA TRIỆT TIÊU NHƯ SAU :
\(-\frac{1}{4}\)RỒI LẠI \(+\frac{1}{4}\)TA TÍNH BÀNG 0.TƯƠNG TỰ CÁC PHÂN SỐ KHÁC TA CŨNG TRIỆT TIÊU ĐI NÊN CHỈ CÒN \(\frac{1}{2}và\frac{1}{128}\)(ĐÓ PHÁ NGOẶC RA BẠN NHÉ.
\(\frac{1}{2}-\frac{1}{128}=\frac{63}{128}\)
A = 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128
A x 2 = 1/4 - ( 1/4 + 1/8 + 1/16 + .............. + 1/64 + 1/128 ) - 1/128
A x 2 = 1/4 - A - 1/128
A x 2 - A = 1/4 - 1/128
A = 1/4 - 1/128
A = 31/128
\(\text{Cách 1: Đặt }\) \(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(\Rightarrow2A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(\Rightarrow2A-A=\frac{1}{2}-\frac{1}{128}\)
\(=>A=\frac{63}{128}\)
\(\text{Cách 2: Ta có:
}\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(=\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+\left(\frac{1}{8}-\frac{1}{16}\right)+\left(\frac{1}{16}-\frac{1}{32}\right)+\left(\frac{1}{32}+\frac{1}{64}\right)\)\(+\left(\frac{1}{64}-\frac{1}{128}\right)\)
\(=\frac{1}{2}-\frac{1}{128}=\frac{63}{128}\)