Đặt \(A=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)

=>\(\frac{1}{3}.A=\frac{1}{3^2}-\frac{2}{3^3}+\frac{3}{3^4}-\frac{4}{3^5}+...+\frac{99}{3^{100}}-\frac{100}{3^{101}}\)

=>\(A+\frac{1}{3}.A=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}+\frac{1}{3^2}-\frac{2}{3^3}+\frac{3}{3^4}-\frac{4}{3^5}+...+\frac{99}{3^{100}}-\frac{100}{3^{101}}\)

=>\(\frac{4}{3}.A=\frac{1}{3}-\left(\frac{2}{3^2}-\frac{1}{3^2}\right)+\left(\frac{3}{3^3}-\frac{2}{3^3}\right)-\left(\frac{4}{3^4}-\frac{3}{3^4}\right)+...+\left(\frac{99}{3^{99}}-\frac{98}{3^{99}}\right)-\left(\frac{100}{3^{100}}-\frac{99}{3^{100}}\right)-\frac{100}{3^{101}}\)

=>\(\frac{4}{3}.A=\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+...+\frac{1}{3^{99}}-\frac{1}{3^{100}}-\frac{100}{3^{101}}\)

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

=>\(\frac{1}{3}.B=\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-\frac{1}{3^5}+...+\frac{1}{3^{100}}-\frac{1}{3^{101}}\)

=>\(B+\frac{1}{3}.B=\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+...+\frac{1}{3^{99}}-\frac{1}{3^{100}}+\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-\frac{1}{3^5}+...+\frac{1}{3^{100}}-\frac{1}{3^{101}}\)

=>\(\frac{4}{3}.B=\frac{1}{3}-\frac{1}{3^{101}}\)

=>\(B=\frac{1}{3}:\frac{4}{3}-\frac{1}{3^{101}}:\frac{4}{3}\)

=>\(B=\frac{1}{3}.\frac{3}{4}-\frac{1}{3^{101}}.\frac{3}{4}\)

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

Lại có: \(\frac{4}{3}.A=\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+...+\frac{1}{3^{99}}-\frac{1}{3^{100}}-\frac{100}{3^{101}}\)

=>\(\frac{4}{3}.A=B-\frac{100}{3^{101}}\)

=>\(\frac{4}{3}.A=\frac{1}{2}-\frac{1}{3^{100}.4}-\frac{100}{3^{101}}\)

=>\(\frac{4}{3}.A=\frac{1}{2}-\left(\frac{1}{3^{100}.4}+\frac{100}{3^{101}}\right)\)

=>\(\frac{4}{3}.A=\frac{1}{2}-\left(\frac{1}{3^{100}}.\frac{1}{4}+\frac{1}{3^{100}}.\frac{100}{3}\right)\)

=>\(\frac{4}{3}.A=\frac{1}{2}-\frac{1}{3^{100}}.\left(\frac{1}{4}+\frac{100}{3^{ }}\right)\)

=>\(\frac{4}{3}.A=\frac{1}{2}-\frac{1}{3^{100}}.\frac{403}{12}\)

Ta thấy: \(\frac{1}{3^{100}}.\frac{403}{12}<\frac{1}{3}.\frac{9}{12}=\frac{1}{3}.\frac{3}{4}=\frac{1}{4}\)

=>\(\frac{1}{3^{100}}.\frac{403}{12}<\frac{1}{4}\)

=>\(\frac{4}{3}.A=\frac{1}{2}-\frac{1}{3^{100}}.\frac{403}{12}<\frac{1}{2}-\frac{1}{4}=\frac{1}{4}\)

=>\(\frac{4}{3}.A<\frac{1}{4}=>A<\frac{1}{4}:\frac{4}{3}=>A<\frac{3}{16}\)

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

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

=))

Dài quá bạn ơi!!!

Mong bạn làm ngắn gọn lại một chút

1/2 ở đâu zậy bạn, phải là 1/4 chứ

làm ngắn lại 1 tí được không

nhìn mà đau đầu quá! *~*

dai dong qua 

Lê Chí Cường làm dài quá chi tiết quá

LÀM NGẮN GỌN ĐƯỢC KO

Đặt \(A=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)

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

\(3A+A=\left(1-\frac{2}{3}+\frac{3}{3^2}-...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\right)+\left(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\right)\)

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

\(12A=3-1+\frac{1}{3}-\frac{1}{3^2}+...-\frac{1}{3^{98}}-\frac{100}{3^{99}}\)

\(4A+12A=\left(1-\frac{1}{3}+\frac{1}{3^2}-...-\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)+\left(3-1+\frac{1}{3}-\frac{1}{3^2}+...-\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)\)

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

\(\Rightarrow16A< 3\)

\(\Rightarrow A< \frac{3}{16}\)(đpcm)

dau ca dau

TXT Channel Funfun có lỗi sai rồi đấy

_{^{qưeqweqeqweqweqw}}

lê chí cường lm sai ở chỗ B-100/3^101  phải là 1/4 ko phải 1/2.

Lê Cường làm sai rồi:  1/4 chứ không phải là 1/2. Thế mà vẫn nhiều thế ~ !

Dài vl ạ

1/3A=1/3^2-2/3^3+3/3^4-4/3^5+...+99/3^100-100/3^101

1/3A+A=4/3A=1/3-(2/3^2-1/3^2)+(3/3^3-2/3^3)+...+(100/3^100-99/3^100)-100/3^101

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

Gọi 1/3-1/3^2+1/3^3-1/3^4+...1/3^99-1/3^100 là B

1/3B=1/3^2-1/3^3+1/3^4-1/3^5+...+1/3^100-1/3^101

1/3B+B=4/3B=1/3-1/3^101

B=1/3:4/3-1/3^101:4/3

B=1/4-1/3^100.4

4/3A=B-100/3^101

4/3A=1/4-1/3^100.4-100/3^101

=> 4/3A<1/4 =>A<1/4:4/3=3/16

=>1/3-2/3^2+3/3^3-4/3^4+...+99/3^99-100/3^100 < 3/16

trả lời ngubv