Đáp án: $A=\dfrac12(9-\dfrac{675}{3^{503}})$
Giải thích các bước giải:
Ta có:
$A=\dfrac73+\dfrac{11}{3^2}+\dfrac{15}{3^3}+....+\dfrac{2019}{3^{504}}$
$\to 3A=7+\dfrac{11}{3}+\dfrac{15}{3^2}+....+\dfrac{2019}{3^{503}}$
$\to 3A-A=7+\dfrac{4}{3}+\dfrac{4}{3^2}+....+\dfrac{4}{3^{503}}-\dfrac{2019}{3^{504}}$
$\to 2A=7+4(\dfrac{1}{3}+\dfrac{1}{3^2}+....+\dfrac{1}{3^{503}})-\dfrac{2019}{3^{504}}$
Mà $B=\dfrac{1}{3}+\dfrac{1}{3^2}+....+\dfrac{1}{3^{503}}$
$\to 3B=1+\dfrac{1}{3}+....+\dfrac{1}{3^{502}}$
$\to 3B-B=1-\dfrac{1}{3^{503}}$
$\to 2B=1-\dfrac{1}{3^{503}}$
$\to B=\dfrac12(1-\dfrac{1}{3^{503}})$
$\to 2A=7+4\cdot \dfrac12(1-\dfrac{1}{3^{503}})-\dfrac{2019}{3^{504}}$
$\to 2A=7+2-\dfrac{2}{3^{503}}-\dfrac{2019}{3^{504}}$
$\to 2A=9-\dfrac{2}{3^{503}}-\dfrac{673}{3^{503}}$
$\to 2A=9-\dfrac{675}{3^{503}}$
$\to A=\dfrac12(9-\dfrac{675}{3^{503}})$
