Giải thích các bước giải:
Ta có :
$B=\dfrac{1}{16}+\dfrac{2}{16^2}+\dfrac{3}{16^3}+..+\dfrac{2018}{16^{2018}}$
$\to 16B=1+\dfrac{2}{16}+\dfrac{3}{16^2}+\dfrac{3}{16^3}+..+\dfrac{2018}{16^{2017}}$
$\to 16B-B=1+\dfrac{1}{16}+\dfrac{1}{16^2}+\dfrac{1}{16^3}+..+\dfrac{1}{16^{2017}}-\dfrac{2018}{16^{2018}}$
$\to 15B=1+\dfrac{1}{16}+\dfrac{1}{16^2}+\dfrac{1}{16^3}+..+\dfrac{1}{16^{2017}}-\dfrac{2018}{16^{2018}}$
Mà $A=1+\dfrac{1}{16}+\dfrac{1}{16^2}+\dfrac{1}{16^3}+..+\dfrac{1}{16^{2017}}$
$\to 16A=16+1+\dfrac{1}{16}+\dfrac{1}{16^2}+..+\dfrac{1}{16^{2016}}$
$\to 16A-A=16-\dfrac{1}{16^{2017}}$
$\to A=\dfrac{16-\dfrac{1}{16^{2017}}}{15}$
$\to 15B=\dfrac{16-\dfrac{1}{16^{2017}}}{15}-\dfrac{2018}{16^{2018}}$
$\to 15B<\dfrac{16}{15}$
$\to B<\dfrac{16}{15^2}<1$
$\to B^{2017}>B^{2018}$
