Question

Answer 1

Ta có:

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

\(\Rightarrow3B=1+\dfrac{2}{3}+\dfrac{3}{3^2}+...+\dfrac{100}{3^{99}}\)

\(\Rightarrow3B-B=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)

\(\Rightarrow2B=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)

\(\Rightarrow6B=3+1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\)

\(\Rightarrow6B-2B=3-\dfrac{101}{3^{99}}+\dfrac{100}{3^{100}}\)

\(\Rightarrow4B=3-\dfrac{101}{3^{99}}+\dfrac{100}{3^{100}}\)

Mà \(\dfrac{100}{3^{99}}>\dfrac{100}{3^{100}}\) nên \(\dfrac{101}{3^{99}}>\dfrac{100}{3^{100}}\)

\(\Rightarrow4B< 3\Rightarrow B< \dfrac{3}{4}\)