Question

Cmr : \(\dfrac{1}{3}\) - \(\dfrac{2}{3^2}\) +\(\dfrac{3}{3^3}\) - \(\dfrac{4}{3^4}\) + ...+\(\dfrac{99}{3^{99}}\) - \(\dfrac{100}{3^{100}}\) < \(\dfrac{3}{16}\)

Answer 1

Answer 2

dễ quá

ha ha ha

ngốc quá đi

Answer 3

thế thì ai làm đi ở dấy mà cười

Answer 4

Answer 5

Sao chỉ có bình luận xuông mà không a quan tâm đến câu hỏi vậy

Ta có :

\(A=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\)

\(\Rightarrow3A=1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\)

\(\Rightarrow4A=3A+A=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)

\(\Rightarrow12A=3.4A=3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...+\dfrac{1}{3^{97}}-\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\)

\(\Rightarrow16A=12A+4A=3-\dfrac{1}{3^{99}}-\dfrac{100}{3^{99}}-\dfrac{100}{3^{100}}\)

\(\Rightarrow16A=3-\dfrac{101}{3^{99}}-\dfrac{100}{3^{100}}\)

\(\Leftrightarrow A=\dfrac{3}{16}-\dfrac{\left(\dfrac{101}{3^{99}}+\dfrac{100}{3^{100}}\right)}{16}< \dfrac{3}{16}\)