\(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+...+\frac{1}{512}-\frac{1}{1024}\)

\(A=\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+...+\frac{1}{2^9}-\frac{1}{2^{10}}\)

\(2A=1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+...+\frac{1}{2^8}-\frac{1}{2^9}\)

\(3A=1-\frac{1}{2^{10}}< 1\)

\(\Rightarrow A< \frac{1}{3}\)

Câu a:

A = 1/2 - 1/4 + 1/8 - 1/16 + 1/32 - 1/64 < 1/3

A = 1/2 - 1/4 + 1/8 - 1/16 + 1/32 - 1/64

2A = 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

3A = (1 - 1/64) + (1/2 - 1/2) + ..+ (1/16 - 1/16)

3A = 1 - 1/64 + 0+.. + 0

3A = 1 - 1/64

A = 1/3 - 1/3.64 < 1/3 (đpcm)

a: Đặt \(A=\frac12-\frac14+\frac18-\frac{1}{16}+\cdots-\frac{1}{1024}\)

=>\(A=\frac12-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\cdots-\frac{1}{2^{10}}\)

=>\(2A=1-\frac12+\frac{1}{2^2}-\frac{1}{2^3}+\cdots-\frac{1}{2^9}\)

=>\(2A+A=1-\frac12+\frac{1}{2^2}-\frac{1}{2^3}+\cdots-\frac{1}{2^9}+\frac12-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\cdots-\frac{1}{2^{10}}\)

=>\(3A=1-\frac{1}{2^{10}}<1\)

=>\(A<\frac13\)

b: Đặt \(B=\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)

=>\(3B=1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\cdots+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)

=>\(3B+B=1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\cdots+\frac{99}{3^{98}}-\frac{100}{3^{99}}+\frac13-\frac{2}{3^2}+\cdots+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)

=>\(4B=1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-...-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

Đặt \(A=-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{99}}\)

=>\(3A=-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{98}}\)

=>\(3A+A=-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{98}}-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)

=>\(4A=-1-\frac{1}{3^{99}}=\frac{-3^{99}-1}{3^{99}}\)

=>\(A=\frac{-3^{99}-1}{4\cdot3^{99}}\)

Ta có: \(4B=1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-...-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

\(=1+\frac{-3^{99}-1}{4\cdot3^{99}}-\frac{100}{3^{100}}=1+\frac{-3^{100}-3-400}{4\cdot3^{100}}=1-\frac14-\frac{403}{4\cdot3^{100}}=\frac34-\frac{403}{4\cdot3^{100}}\)

=>\(4B<\frac34\)

=>\(B<\frac{3}{16}\)

ai nhanh mình k

1 /2 -1 /4 + 1 /8-1 /16 + 1 /32-1 /64 < 1 /3

Cách 1:21/64 < 1/3

Cách 2:21/64 < 0.(3)

Đúng

1 /2 + 1 /4 + 1 /8 + 1 /16 + 1 /32 + 1 /64 < 1 /3

Cách 2:63/64 < 0.(3)

Ko đúng

Câu 3 mình ko biết

Ai giúp cái, tui k cho

cm A<1/2

và B<3/2 thì có thể nhưng bất đẳng thức thì ko có đâu