Giải thích các bước giải:
Đặt:
$A=\dfrac13-\dfrac2{3^2}+\dfrac3{3^3}-...-\dfrac3{3^{100}}$
$\to 3A=1-\dfrac2{3}+\dfrac3{3^2}-...-\dfrac3{3^{99}}$
$\to 3A+A=1-\dfrac13+\dfrac1{3^2}-\dfrac1{3^3}+...-\dfrac{1}{3^{99}}-\dfrac3{3^{100}}$
$\to 4A=1-\dfrac13+\dfrac1{3^2}-\dfrac1{3^3}+...-\dfrac{1}{3^{99}}-\dfrac3{3^{100}}$
Ta có:
$B=1-\dfrac13+\dfrac1{3^2}-\dfrac1{3^3}+...-\dfrac{1}{3^{99}}$
$\to 3B=3-1+\dfrac1{3}-\dfrac1{3^2}+...-\dfrac{1}{3^{98}}$
$\to 3B+B=3-\dfrac1{3^{99}}$
$\to 4B=3-\dfrac1{3^{99}}$
$\to B=\dfrac34- \dfrac1{4\cdot 3^{99}}$
$\to 4A=\dfrac34-\dfrac1{4\cdot 3^{99}}-\dfrac3{3^{100}}$
$\to 4A<\dfrac34$
$\to A<\dfrac3{16}$
