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Câu hỏi của Ngô Văn Nam - Chuyên mục hỏi đáp - Giúp tôi giải toán. - Học toán với OnlineMath

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

\(\Rightarrow3A=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)

\(\Rightarrow3A+A=1+\left(\frac{1}{3}-\frac{2}{3}\right)+\left(\frac{-2}{3^2}+\frac{3}{3^2}\right)+\left(\frac{3}{3^3}-\frac{4}{3^3}\right)+...+\left(\frac{-98}{3^{98}}+\frac{99}{3^{98}}\right)+\left(\frac{99}{3^{99}}-\frac{100}{3^{99}}\right)-\frac{100}{3^{100}}\)

\(\Rightarrow4A=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

\(\Rightarrow3.4A=3-1+\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}-\frac{100}{3^{99}}\)

\(\Rightarrow3.4A+4A=3+\left(1-1\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{3^2}-\frac{1}{3^2}\right)+...+\left(\frac{1}{3^{98}}-\frac{1}{3^{98}}\right)-\frac{101}{3^{99}}-\frac{100}{3^{100}}\)

\(\Rightarrow16A=3-\frac{99}{3^{99}}-\frac{100}{3^{100}}< 3\Rightarrow A< \frac{3}{16}< \frac{3}{4}\)

Ta có:
A=1/3 - 2/3^2+3/3^3 - 4/3^4+ ... - 100/3^100
=>3A=1 -2/3 +3/3^2 - 4/3^3+ ... - 100/3^99
=>4A=A+3A=1-1/3+1/3^2-1/3^3+...-1/3^99 - 100/3^100
=>12A=3.4A=3-1+1/3-1/3^2+...-1/3^98 - 100/3^99

=>16A=12A+4A=3-1/3^99-100/3^99-100/3^1...
<=>16A=3-101/3^99-100/3^100
<=>A=3/16-(101/3^99+100/3^100)/16 < 3/16
Suy ra A<3/16

ai tk mình đi đang bị âm điểm nè

cảm ơn các bạn nhìu!!!

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 \(A<\frac14\)

Câu hỏi của Biêtdongsaigon - Toán lớp 6 - Học toán với OnlineMath

Bạn tham khảo link này nhé!

k cho tôi đấy nhá An

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

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

=>4A= 1 - 1/3 + 1/3^2 - 1/3^3 +...+ 1/3^98 - 1/3^99 - 100/3^100

=>4A<1 - 1/3 + 1/3^2 - 1/3^3 +...+ 1/3^98 -1/3^99

=>4A<1-(1/3 -1/3^2+1/3^3-...-1/3^98+1/3^99)

Đặt B=1/3 -1/3^2+1/3^3-...-1/3^98+1/3^99

=>3B=1 - 1/3 +1/3^2 -... - 1/3^97 +1/3^98

=>4B=1+1/3^99>1

=>4B>1

=>B>1/4

=>-B<-1/4

=>1-B<1-1/4

=>4A<1-B<3/4

=>4A<3/4

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

=>A<3/16 (đpcm)