dốt thế 

Mình ngu lắm dân trần đăng ninh chuyên anh mà làm sao giỏi toán được

\(Cm:\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)

Gọi biểu thức trên là A, ta có:

3A = 1-2/3+3/3^2-...-100/3^99

3A + A = [1-2/3+3/3^2-...-100/3^99] + [1/3-2/3^2+3/3^3-...-100/3^100]

4A = 1 - 1/3 + 1/3^2 - ... - 1/3^99 - 100/3^99 [1]

Gọi B = 1-1/3 + 1/3^2 - ... - 1/3^99

3B = 3 - 1 + 1/3 - 1/3^2 -...-1/3^2012

3B + B = [3-1+1/3-1/3^2-...-1/3^2012] + [1-1/3 + 1/3^2 - ... - 1/3^99]

4B = 3 - 1/3^99 

=> 4B < 3 => B < 1/4 [2]

Từ [1], [2] => 4A < B < 3/4 => A < 3/16 [đpcm]

MỎI TAY QUỚ

tk nha

Lúc đặt câu hỏi, bạn bấm vào góc trên cùng bên trái để gõ phép tính đẹp. Ý của bạn có phải là:

\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)

gọi biểu thức trên là A , ta có :

\(A=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+\dfrac{5}{3^5}-...+\dfrac{99}{3^{99}}+\dfrac{100}{3^{100}}\\ 3A=1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\\ \Rightarrow A+3A=\left(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\right)+\left(1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\right)\\ \Rightarrow4A\cdot3=12A=3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}\)

từ đó ta được :

\(16A=3-\dfrac{100}{3^{99}}-\dfrac{100}{3^{100}}\\ \Rightarrow A=\dfrac{\dfrac{3-101}{3^{99}}-\dfrac{100}{3^{100}}}{16}\\ \Rightarrow A=\dfrac{3}{16}-\dfrac{\dfrac{101}{3^{99}}-\dfrac{100}{3^{100}}}{16}< \dfrac{3}{16}\)

help mik với 

lx

ko có câu hỏi bn ơi

Sửa đề: Chứng minh B<1

Ta có: \(B=\frac13+\frac{2}{3^2}+\cdots+\frac{100}{3^{100}}\)

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

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

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

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

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

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

=>\(2A=1-\frac{1}{3^{99}}=\frac{3^{99}-1}{3^{^{99}}}\)

=>\(A=\frac{3^{99}-1}{2\cdot3^{99}}\)

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

\(=1+\frac{3^{99}-1}{2\cdot3^{99}}-\frac{100}{3^{100}}=1+\frac{3^{100}-3-200}{2\cdot3^{100}}=1+\frac12-\frac{203}{2\cdot3^{100}}\) <3/2

=>B<3/4

Ta có: \(S=\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots-\frac{100}{3^{100}}\)

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

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

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

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

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

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

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

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

=>S<3/16

mà 3/16<3/15=1/5

nên S<1/5

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

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

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

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

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

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

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

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

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

Ta có: \(4A=1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

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

=>\(4A<\frac34\)

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

mà \(\frac{3}{16}<\frac{4}{16}=\frac14\)

nên \(A<\frac14\)