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)