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Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)\)
\(A=1-\frac{1}{64}\)
\(A=\frac{63}{64}\)
=32/64+14/64+8/64+4/64+2/64+1/64
=32+14+8+4+2+1/64
=61/64
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}+\frac{1}{32}-\frac{1}{64}\)
\(=1-\frac{1}{64}\)
\(=\frac{63}{64}\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
đặt A = \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(A\cdot2=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(A\cdot2-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)\) \(-\) \(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\right)\)
\(A=\) \(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-\frac{1}{32}-\frac{1}{64}-\frac{1}{128}\)
\(A=1-\frac{1}{128}\)
\(A=\frac{127}{128}\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
Ta lấy\(\frac{1}{128}\)là MSC. Ta tính được \(\frac{64}{128}+\frac{32}{128}+\frac{16}{128}+\frac{8}{128}+\frac{4}{128}+\frac{2}{128}+\frac{1}{128}\)
Kết quả bằng \(\frac{127}{128}\)
1/2 + 1/4 + 1/8 + 1/16 +1/32 + 1/64 + 1/128
=1-1/2+1/2-1/4+1/4-1/8+...+1/64+1/128
=1-1/128
=127/128
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{64}-\frac{1}{128}\)
\(=1-\frac{1}{128}\)
\(=\frac{127}{128}\)
A=\(\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)=\frac{1}{2}\left(1+A-\frac{1}{2}-\frac{1}{128}\right)\)
2A=\(A+\frac{1}{2}-\frac{1}{128}=A+\frac{63}{128}\)
=> A=\(\frac{63}{128}\)
Đặt \(A=1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-..-\frac{1}{2048}\)
\(\Rightarrow A=1-\left(1-\frac{1}{2}\right)-\left(\frac{1}{2}-\frac{1}{4}\right)-..-\left(\frac{1}{1024}-\frac{1}{2048}\right)\)
\(\Rightarrow A=1-1+\frac{1}{2}-\frac{1}{2}+\frac{1}{4}-..-\frac{1}{1024}+\frac{1}{2018}\)
\(\Rightarrow A+\frac{1}{2018}\)
1-1/2-1/4-1/8-1/16-1/32-1/64-1/128-1/256-1/512-1/1024-1/2048 =0.00048828125
Đặt biểu thức trên bằng A
=> \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
=> \(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)
=> A= 2A-A = 1- 1/32= 31/32
Vậy A= 31/32
b: A=1/3+1/9+...+1/3^10
=>3A=1+1/3+...+1/3^9
=>A*2=1-1/3^10=(3^10-1)/3^10
=>A=(3^10-1)/(2*3^10)
c: C=3/2+3/8+3/32+3/128+3/512
=>4C=6+3/2+...+3/128
=>3C=6-3/512
=>C=1023/512
d: A=1/2+...+1/256
=>2A=1+1/2+...+1/128
=>A=1-1/256=255/256
1/2 + 1/4+ 1/8+ 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512
= 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 + 1/128 – 1/256 – 1/256 – 1/512
= 1 – 1/512
= 511/512 .