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

1³+2³+3³+...+100³

=(1+2+3+4+5+...+100)³

=5050³

=128787625000

\(3+3^2+3^3+3^4+\cdots+3^{99}+3^{100}\)

\(=\left(3+3^2\right)+\left(3^3+3^4\right)+\cdots+\left(3^{99}+3^{100}\right)\)

\(=3\left(1+3\right)+3^3\left(1+3\right)+\cdots+3^{99}\left(1+3\right)\)

\(=4\left(3+3^3+\cdots+3^{99}\right)\) ⋮4

2

x=2