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đo jgfhjgh dfj hdfh vhfcvkjgb 

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 

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

A=-1+(-1)+...+(-1) {có 50 số hạng}

=-1.50=-50

A=1-2+3-4+...+99-100       SSH=(100-1):1+1=100 Sh

=>A=(1-2)+(3-4)+....+(99-100)

vì chia thành cặp suy ra 100:2 =50 cặp

A=(-1)+(-1)+...(-1)

A=(-1).50

A=-50

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