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1) 1/1.2 + 1/2.3 + ... + 1/6.7
= 1 - 1/2 + 1/2 - 1/3 + ... + 1/6 - 1/7
= 1 - 1/7
= 6/7
2) 1/2 + 1/6 + 1/12 + .. + 1/72
= 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/8.9
= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/8 - 1/9
= 1 - 1/9
= 8/9
3) \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2019}\right)\)
= \(\frac{1}{2}.\frac{2}{3}...\frac{2019}{2020}\)
= \(\frac{1.2....2019}{2.3...2020}\)
= \(\frac{1}{2020}\)
4) A = \(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{512}\)
= \(\frac{1}{2^2}+\frac{2}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^9}\)
=> 2A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}\)
Lấy 2A - A = \(\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}\right)-\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^9}\right)\)
A = \(\frac{1}{2}-\frac{1}{2^9}\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
\(< =>\frac{128}{256}+\frac{64}{256}+\frac{32}{256}+\frac{16}{256}+\frac{8}{256}+\frac{4}{256}+\frac{2}{256}+\frac{1}{256}\)
\(< =>\frac{128+64+32+16+8+4+2+1}{256}\)
\(< =>\frac{255}{256}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(< =>\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(< =>\frac{1}{1}-\frac{1}{100}\)
\(< =>\frac{99}{100}\)
\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{100}\right)\)
\(< =>\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{99}{100}\)
\(< =>\frac{1\cdot2\cdot3\cdot...\cdot99}{2\cdot3\cdot4\cdot...\cdot100}\)
\(< =>\frac{1}{100}\)
mk chuc ban hoc tot nhe :))
Đặt A = 1/2 − 1/4 + 1/8 − 1/16 + 1/32 − 1/64
A = 1/2 − 1/4 + 1/8 − 1/16 + 1/32 − 1/64
2A = 1 − 1/2 + 1/4 − 1/8 + 1/16 − 1/322
A =1 − 1/2 + 1/4 − 1/8 + 1/16 − 1/32
3A = 2A + A = 1 − 1/64 < 1
⇒ A < 1/3
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CHÚC BẠN HỌC TỐT!!!!!
Tk cho mình nha
Ta có : \(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{64}\)
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^6}\)
\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^5}\)
\(\Rightarrow2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^5}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^6}\right)\)
\(\Rightarrow A=1-\dfrac{1}{2^6}=1-\dfrac{1}{64}=\dfrac{63}{64}\)
\(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{8}+...+\dfrac{1}{32}-\dfrac{1}{64}\)
\(=1-\dfrac{1}{64}\)
\(=\dfrac{63}{64}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{32}-\frac{1}{64}\)
\(=\frac{1}{1}-\frac{1}{64}=\frac{63}{64}\)
a) \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{64}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{32}\)
\(\Rightarrow2A-A=A=1-\frac{1}{64}=\frac{63}{64}\)
A=1/2 - 1/4 + 1/8 - 1/16 + 1/32 - 1/64
A= ( 1/2 - 1/4 ) + ( 1/8 - 1/16 ) + ( 1/32 - 1/64 )
A= 1/4 + 1/16 + 1/64
A = 16/64 + 4/64 + 1/64
A = 16+4+1/64
A= 21/64
Ta có : 1/3 = 21/63 mà 21/64 < 21/63 => 21/64 < 1/3 => 1/2 - 1/4 + 1/8 - 1/16 + 1/32 - 1/ 64 < 1/3
Vậy 1/2 - 1/4 + 1/8 - 1/16 + 1/32 - 1/ 64 < 1/3 ( đã chứng minh được )
S là số vô hạn thì điều đó đúng. Còn S không phải là số vô hạn thì điều đó sai.
2s = 2+4 +.......128 +..... chứ k phai 64, bạn khôn quá he
nên 2s khác s-1 nghe bạn , k lừa dc tui đâu