Giải thích các bước giải:
Ta có :
$A=\dfrac{1}{3}+\dfrac{4}{3^2}+\dfrac{7}{3^3}+..+\dfrac{3n-2}{3^n}$
$\to 3A=1+\dfrac{4}{3}+\dfrac{7}{3^2}+..+\dfrac{3n-2}{3^{n-1}}$
$\to 3A-A=1+\dfrac{4-1}{3}+\dfrac{7-4}{3^2}+..+\dfrac{3n-2-(3(n-1)-2)}{3^{n-1}}+\dfrac{3n-2}{3^{n-1}}$
$\to 2A=1+\dfrac{3}{3}+\dfrac{3}{3^2}+..+\dfrac{3}{3^{n-1}}+\dfrac{3n-2}{3^{n-1}}$
$\to 2A=1+1+\dfrac{1}{3}+..+\dfrac{1}{3^{n-2}}+\dfrac{3n-2}{3^{n-1}}$
$\to 2A=1+(1+\dfrac{1}{3}+..+\dfrac{1}{3^{n-2}})+\dfrac{3n-2}{3^{n-1}}$
$\to 2A=1+\dfrac{\dfrac{1}{3^{n-1}}-1}{\dfrac{1}{3}-1}+\dfrac{3n-2}{3^{n-1}}$
$\to 2A=1-\dfrac{\dfrac{1}{3^{n-1}}-1}{\dfrac{2}{3}}++\dfrac{3n-2}{3^{n-1}}$
$\to 2A=1-\dfrac{1}{2.3^{n}}+\dfrac{3}{2}+\dfrac{3n-2}{3^{n-1}}$
$\to 2A=\dfrac{5}{2}-\dfrac{1}{2.3^{n}}+\dfrac{3n-2}{3^{n-1}}$
$\to A=\dfrac{5}{4}-\dfrac{1}{4.3^{n}}+\dfrac{3n-2}{2.3^{n-1}}$
