@Đỗ Nguyễn Như Bình \(\frac{2}{3^2}\) hay là \(\frac{2^2}{3}\) hay là \(\left(\frac{2}{3}\right)^2\) vậy em???????????

1/3E=1/3^2+2/3^3+...+100/3^101

E-1/3E=1/3+1/3^2+1/3^3+...+1/3^100-1/3^101

2/3E=1/3+1/3^2+1/3^3+...+1/3^100-1/3^101

Đặt B=1/3+1/3^2+...+1/3^100

      1/3B=1/3^2+1/3^3+...+1/3^101

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

2/3B=1/3-1/3^101

mà 1/3-1/3^101<1/3

=>2/3B<1/3

=>B<1/2

thay B vào E ta có

2/3E=B-1/3^101

Mà B-1/3^101<B

=>2/3E<B

Mà B<1/2

=>2/3E<1/2

=>E<3/4

k cho mk nha

ta có : 1+1+1+1+1+1+1+1x0

=> 1x8 = 8

mà kòn x vs 0 nữa :

=> tổng đó =0

=> 0<3/4

=> E<3/4

$C=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}$

$3C=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}$

$3C-C=\left(1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\right)$

$2C=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}$

$6C=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}$

$6C-2C=\left(3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)$

$4C=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}$

$4C=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}$

$4C=3-\frac{203}{3^{100}}<3$

$\Rightarrow C<\frac{3}{4}\left(đpcm\right)$

k nha

$C=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}$

$3C=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}$

$3C-C=\left(1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\right)$

$2C=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}$

$6C=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}$

$6C-2C=\left(3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)$

$4C=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}$

$4C=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}$

$4C=3-\frac{203}{3^{100}}<3$

$\Rightarrow C<\frac{3}{4}\left(đpcm\right)$