Câu 2 : Tính nhanh:
\(\dfrac{1}{2}\)+\(\dfrac{1}{4}\)+\(\dfrac{1}{8}\)+...+\(\dfrac{1}{256}\)+\(\dfrac{1}{512}\)
S=\(\dfrac{1}{2}\)+\(\dfrac{1}{4}\)+\(\dfrac{1}{8}\)+\(\dfrac{1}{16}\)+\(\dfrac{1}{32}\)+\(\dfrac{1}{64}\)+\(\dfrac{1}{128}\)+\(\dfrac{1}{256}\)+\(\dfrac{1}{512}\)+\(\dfrac{1}{1024}\)
tính nhanh \(\dfrac{1}{2}\) +\(\dfrac{1}{4}\) +\(\dfrac{1}{8}\)+\(\dfrac{1}{16}\) +.....+\(\dfrac{1}{512}\) +\(\dfrac{1}{1024}\)
Đặt A=1/2+1/4+1/8+..+1/1024
Ax2=1+1/2+1/4+1/8+..+1/512( Nhân cả 2 vế với 2)
Ax2-A=(1+1/2+1/4+1/8+..+1/512)-(1/2+1/4+1/8+..+1/1024)
<=>A=1-1/1024
<=>A=1023/1024
Vậy biểu thức đã cho = 1023/1024
\(\dfrac{1}{2}\)+\(\dfrac{1}{4}\)+\(\dfrac{1}{8}\)+\(\dfrac{1}{16}\)+\(\dfrac{1}{32}\)+\(\dfrac{1}{64}\)+\(\dfrac{1}{128}\)+\(\dfrac{1}{256}\)
Tính nhanh
\(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) +............ + \(\dfrac{1}{256}\) + \(\dfrac{1}{512}\) = ?
\(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+..........+\dfrac{1}{256}+\dfrac{1}{512}=?\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{8}+...+\dfrac{1}{256}-\dfrac{1}{512}-\dfrac{1}{512}\)
\(=1-\dfrac{1}{512}\)
\(=\dfrac{511}{512}\)
Vậy \(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+.........+\dfrac{1}{256}+\dfrac{1}{512}=\dfrac{511}{512}\)
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Đặt :
\(S=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...........+\dfrac{1}{256}+\dfrac{1}{512}\)
\(\Leftrightarrow S=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+.........+\dfrac{1}{2^8}+\dfrac{1}{2^9}\)
\(\Leftrightarrow2S=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+.........+\dfrac{1}{2^9}\right)\)
\(\Leftrightarrow2S=1+\dfrac{1}{2}+\dfrac{1}{2^2}+.........+\dfrac{1}{2^8}\)
\(\Leftrightarrow2S-S=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^8}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+......+\dfrac{1}{2^9}\right)\)
\(\Leftrightarrow S=1-\dfrac{1}{2^9}\)
\(\Leftrightarrow S=1-\dfrac{1}{512}=\dfrac{511}{512}\)
Ta có : \(\dfrac{1}{2}\) = 1 - \(\dfrac{1}{2}\) ; \(\dfrac{1}{4}\) = \(\dfrac{1}{2}\) - \(\dfrac{1}{4}\) ; \(\dfrac{1}{8}\) = \(\dfrac{1}{4}\) - \(\dfrac{1}{8}\) ;............. ;
\(\dfrac{1}{512}\) = \(\dfrac{1}{256}\) - \(\dfrac{1}{512}\)
Vậy : \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + ................... + \(\dfrac{1}{256}\) + \(\dfrac{1}{512}\)
= 1 - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{8}\) + ............. + \(\dfrac{1}{256}\) - \(\dfrac{1}{512}\)
= 1 - \(\dfrac{1}{512}\)
= \(\dfrac{511}{512}\)
A=\(\dfrac{1}{2}\)+\(\dfrac{1}{4}\)+\(\dfrac{1}{8}\)+....+\(\dfrac{1}{512}\)+\(\dfrac{1}{1024}\)
\(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{512}+\dfrac{1}{1024}\)
\(=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\)
\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)
\(\Rightarrow2A-A=A=1-\dfrac{1}{2^{10}}\)
Cho \(A=1-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{8}-\dfrac{1}{10}-...-\dfrac{1}{512}.CM:A< 0,002\)
Lời giải:
Sửa lại đề:
\(A=1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-...-\frac{1}{512}\)
\(=1-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\right)\)
\(2A=2-\left(1+\frac{1}{2}+...+\frac{1}{2^8}\right)\)
Trừ theo vế:
\(A=2A-A=\frac{1}{2^9}< 0,002\) (đpcm)
chứng minh rằng
a , \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+...+\dfrac{1}{512}-\dfrac{1}{1024}\) < \(\dfrac{1}{3}\)
b , \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\) < \(\dfrac{3}{16}\)
(2 + 4 + 6 + ... + 98) x (\(\dfrac{3}{4}\) x 8 - 15 x 0,4) x (\(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + ... + \(\dfrac{1}{512}\) )
Giúp mình vớiii
\(=\left(2+4+6+...+98\right)\left(6-6\right)\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{512}\right)\)
=0
= ( 2+4+6+...+98 ) ( 6- 6) ( 1/2+1/4 + .......+ 1/ 512 )
= 0
Chúc bạn học tốt
= (2 + 4 + 6 + ... + 98) x ( 6 - 6 ) x ( \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) +...+ \(\dfrac{1}{512}\))
= (2 + 4 + 6 + ... + 98) x 0 x ( \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) +...+ \(\dfrac{1}{512}\))
= 0
*****Vì số nào nhân với 0 cũng bằng 0
A = \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\)+ \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\) + \(\dfrac{1}{64}\) + \(\dfrac{1}{128}\) + \(\dfrac{1}{256}\)
\(A=\left(1-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{128}-\dfrac{1}{256}\right)\)
\(A=1-\dfrac{1}{256}\)
\(A=\dfrac{255}{256}\)