cmr
a) 1/2 -1/4+1/8-1/16+1/32-1/64 <1/3
b) 1/3-2/3^2+3/3^3-4/3^4+...+99/3^99-100/3^100<3/16
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1/2+1/4+1/8+1/16+1/32+1/64
= 2 x (1/2+1/4+1/8+1/16+1/32+1/64)
= 1 + 1/2+1/4+1/8+1/16+1/32
=> 2A - A = (1+1/2+1/4+1/8+1/16+1/32) - (1/2+1/4+1/8+1/16+1/32+1/64)
=> A = 1 - 1/64
= 63/64
\(2xB=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}\)
\(B=2xB-B=1-\dfrac{1}{64}=\dfrac{63}{64}\)
1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64
= 1 – 1/2 + 1/2 - 1/4 + 1/4 - 1/8 + 1/8 - 1/16 + 1/16 - 1/32 + 1/32 - 1/64
= 1 – 1/64
= 63/64
Bên trên mik trình bày như vậy cho bạn dễ nhìn nha!
\(A=\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-\frac{1}{32}-\frac{1}{64}\)
\(\Leftrightarrow\)\(2A=1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-\frac{1}{32}\)
\(\Leftrightarrow\)\(2A-A=\left(1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-\frac{1}{32}\right)\)\(-\left(\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-\frac{1}{8}-\frac{1}{16}-\frac{1}{32}-\frac{1}{64}\right)\)
\(\Leftrightarrow\)\(A=1-\frac{1}{64}\)\(=\frac{63}{64}\)
1/2 - 1/4 - 1/8 - 1/16 - 1/32 - 1/64
= 32/64 - 16/64 - 8/64 - 4/64 - 2/64 - 1/64
= 1/64 .
^ - ^ . Mình không chăc chắn lắm đâu !
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{32}+\frac{1}{64}\)
=\(\frac{1}{1\cdot2}+\frac{1}{2\cdot2}+\frac{1}{2\cdot4}+\frac{1}{2\cdot16}+\frac{1}{2\cdot32}\)
=\(\frac{1}{2}\cdot\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{16}+\frac{1}{32}\right)\)
=\(\frac{1}{2}\cdot\frac{119}{64}\)
=\(\frac{119}{128}\)