Question

Tính \(D=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+.....+\dfrac{1}{3^{100}}\)

Answer 1

$D=\frac{1}{3}+\frac{1}{{}^{}}+\frac{1}{{}^{}}+...+\frac{1}{{}^{}}$

$\iff 3D=1+\frac{1}{3}+\frac{1}{{}^{}}+...+\frac{{}^{}}{}$

$\iff 3D-D=(1+\frac{1}{3}+\frac{1}{{}^{}}+...+\frac{1}{{}^{}})-(\frac{1}{3}+\frac{1}{{}^{}}+\frac{1}{{}^{}}+...+\frac{1}{{}^{}})$

$\iff 2D=1+\frac{1}{}$

Answer 2

\(D=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\)

\(3D=3+1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)

\(3D-D=3+1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}-1-\dfrac{1}{3}-\dfrac{1}{3^2}-\dfrac{1}{3^3}-...-\dfrac{1}{3^{100}}\)

\(2D=3-\dfrac{1}{3^{100}}\)

\(2D=\dfrac{3^{111}-1}{3^{100}}\)

\(D=\dfrac{3^{111}-1}{2.3^{100}}\)