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\(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+....+\frac{1}{256}-\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{512}\)
\(=\frac{255}{512}\)
Vậy \(A=\frac{255}{512}\)
A=14 +18 +116 +132 +164 +1128 +1256 +1512
=12 −14 +14 −18 +....+1256 −1512
=12 −1512
=255512
Vậy A=255512
Phạm Long Khánh
\(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-....+\frac{1}{256}-\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{512}\)
\(=\frac{255}{512}\)
Vậy \(A=\frac{255}{512}\)
=1/2-1/4+1/4-1/8+1/8-....+1/156-1/152
=1/2-1/152
=255/512
A=255/512
Đặt A = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512
2A = 1/2 x 2 + 1/4 x 2 + 1/8 x 2 + 1/16 x 2 + 1/32 x 2 + 1/64 x 2 + 1/128 x 2 + 1/256 x 2 + 1/512 x 2
2A = 1 + 1/2 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256
2A - A = ( 1 + 1/2 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256 ) - ( 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512 )
A = 1 - 1/512
A = 511/512
Ta có : \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\)
Đặ A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\)(1)
=> 2A = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}+\frac{1}{2^9}\)(2)
Lấy (2) trừ (1) theo vế ta có :
2A - A = \(\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\right)\)
=> A = \(1-\frac{1}{2^{10}}=\frac{2^{10}-1}{2^{20}}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{10}}\)
\(\Leftrightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^9}\)
\(\Rightarrow2A-A=1-\frac{1}{2^{10}}=\frac{1023}{1024}\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}+\frac{1}{32}-\frac{1}{64}+\frac{1}{64}-\frac{1}{128}\)
\(=1-\frac{1}{128}\)
\(\frac{127}{128}\)
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Câu 2:
2Q = \(\frac{2}{1.2.3}\) + \(\frac{2}{2.3.4}\) + ... + \(\frac{2}{98.99.100}\)
2Q = \(\frac12\).(\(\frac{2}{1.3}\)) + \(\frac13\).(\(\frac{2}{2.4}\)) + ... + \(\frac{1}{99}\).(\(\frac{2}{98.100}\))
2Q = \(\frac12\).(\(\frac11-\frac13\)) + \(\frac13\).(\(\frac12-\frac14\)) + ...+ \(\frac{1}{99}\).(\(\frac{1}{98}-\frac{1}{100}\))
2Q = \(\frac{1}{1.2}\) - \(\frac{1}{2.3}\) + \(\frac{1}{2.3}\) - \(\frac{1}{3.4}\) + ...+\(\frac{1}{98.99}\) - \(\frac{1}{99.100}\)
2Q = \(\frac12-\frac{1}{9900}\)
2Q = \(\frac{4949}{9900}\)
Q = \(\frac{4949}{9900}\) : 2
Q = \(\frac{4949}{19800}\)
Câu 1:
A = \(\frac14+\frac18+\frac{1}{16}+..+\frac{1}{128}\)
2A = \(\frac12+\frac14+\frac18+\cdots+\frac{1}{64}\)
2A - A = \(\frac12+\frac14+\frac18+\cdots+\frac{1}{64}\) - \(\frac14-\frac15-\frac{1}{16}-\ldots\frac{1}{128}\)
A = (\(\frac12-\frac{1}{128})+\left(\frac14-\frac14)+..+\left(\frac{1}{64}-\frac{1}{64}\right)\right.\)
A = \(\frac{64}{128}-\frac{1}{128}\) + 0 + 0+..+0
A = \(\frac{63}{128}\)
\(2A=2\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)
\(2A=1+\frac{1}{2}+...+\frac{1}{2^9}\)
\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)
\(A=1-\frac{1}{2^{10}}=\frac{2^{10}-1}{2^{10}}=\frac{1023}{1024}\)
BẤM ĐÚNG NHÉ
1023/1024 nhé bạn
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\)
\(2A=\frac{1}{2}\times2+\frac{1}{4}\times2+\frac{1}{8}\times2+...+\frac{1}{512}\times2+\frac{1}{1024}\times2\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}+\frac{1}{512}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}+\frac{1}{512}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\right)\)
\(A=1-\frac{1}{1024}\)
\(A=\frac{1023}{1024}\)