Câu hỏi của Hoàng Đức Tùng

3S=1-2/3+3/3^2-...+99/3^98-100/3^99=>4S=1-1/3+1/3^2-1/3^3+...+1/3^99-100/3^100=>12S=3-1+1/3-1/3^2+...+1/3^98-100/3^99=>16S=3-99/3^99-100/3^100=>\(S=\dfrac{3}{16}-\dfrac{\dfrac{99}{3^{99}}+\dfrac{100}{3^{100}}}{16} 4S=1-1/3+1/3^2-1/3^3+...+1/3^99-100/3^100=>12S=3-1+1/3-1/3^2+...+1/3^98-100/3^99=>16S=3-99/3^99-100/3^100=>\(S=\dfrac{3}{16}-\dfrac{\dfrac{99}{3^{99}}+\dfrac{100}{3^{100}}}{16}< \dfrac{3}{15}=\dfrac{1}{5}\)