1) Tính tổng C = \(\left(1-\dfrac{1}{2}\right)\).\(\left(1-\dfrac{1}{3}\right)\).\(\left(1-\dfrac{1}{4}\right)\).....\(\left(1-\dfrac{1}{2022}\right)\)
2) Cho tổng A = \(\dfrac{1}{3}\) - \(\dfrac{2}{3^2}\) + \(\dfrac{3}{3^3}\) - \(\dfrac{4}{3^4}\) +...+ \(\dfrac{99}{3^{99}}\) - \(\dfrac{100}{3^{100}}\). Chứng tỏ rằng A...
Đọc tiếp
1) Tính tổng C = \(\left(1-\dfrac{1}{2}\right)\).\(\left(1-\dfrac{1}{3}\right)\).\(\left(1-\dfrac{1}{4}\right)\).....\(\left(1-\dfrac{1}{2022}\right)\)
2) Cho tổng A = \(\dfrac{1}{3}\) - \(\dfrac{2}{3^2}\) + \(\dfrac{3}{3^3}\) - \(\dfrac{4}{3^4}\) +...+ \(\dfrac{99}{3^{99}}\) - \(\dfrac{100}{3^{100}}\). Chứng tỏ rằng A < \(\dfrac{3}{16}\)
1) Ta có
\(C=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2022}\right)\)
\(C=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2021}{2022}\)
\(C=\dfrac{1}{2022}\)
2) \(A=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\)
\(\Rightarrow3A=1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\)
\(\Rightarrow4A=A+3A\) \(=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)
\(\Rightarrow12A=3.4A=3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...-\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\)
\(\Rightarrow16A=12A+4A=\left(3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...-\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\right)+\left(1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\right)\)
\(=3-\dfrac{101}{3^{99}}-\dfrac{100}{3^{100}}\) \(< 3\). Từ đó suy ra \(A< \dfrac{3}{16}\)