Violympic toán 7
Các câu hỏi tương tự
Chứng minh rằng : \(\dfrac{1}{6}< \dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+....+\dfrac{1}{100^2}< \dfrac{1}{4}\)
Chứng minh rằng : \(\dfrac{1}{6}< \dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}\)
Tính tổng đại số
\(A=\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{2}{4}+\dfrac{3}{4}-\dfrac{1}{5}-\dfrac{2}{5}-\dfrac{3}{5}-\dfrac{4}{5}+...+\dfrac{1}{10}+\dfrac{2}{10}+...+\dfrac{9}{10}\)
\(B=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{2}{4}+\dfrac{3}{4}+...+\dfrac{1}{n}+\dfrac{2}{n}+...+\dfrac{n-1}{n}\)\(\left(n\in Z,n\ge2\right)\)
Cho S=\(\dfrac{1}{5^2}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+\dfrac{4}{5^4}+...+\dfrac{2017}{5^{2017}}+\dfrac{2018}{5^{2018}}\).Chứng minh S<\(\dfrac{1}{3}\)
Chứng minh rằng: N = \(\dfrac{1}{5}\)+\(\dfrac{1}{5^2}\)+\(\dfrac{1}{5^3}\)+...+\(\dfrac{1}{5^{99}}\)<\(\dfrac{1}{4}\)
Rút gọn ;
D = \(\dfrac{1}{5}-\dfrac{1}{5^2}+\dfrac{1}{5^3}-\dfrac{1}{5^4}+\dfrac{1}{5^5}-\dfrac{1}{5^6}+...+\dfrac{1}{5^{99}}-\dfrac{1}{5^{100}}+\dfrac{1}{6.5^{100}}\)
2 tháng 7 2021 lúc 21:10
Tính giá trị biểu thức A , biết rằng A M : N Mà M dfrac{dfrac{1}{99}+dfrac{2}{98}+dfrac{3}{97}+dfrac{4}{96}+...+dfrac{97}{3}+dfrac{98}{2}+dfrac{99}{1}}{dfrac{1}{2}+dfrac{1}{3}+dfrac{1}{4}+dfrac{1}{5}+dfrac{1}{6}+...+dfrac{1}{100}} N dfrac{92-dfrac{1}{9}-dfrac{2}{10}-dfrac{3}{11}-...-dfrac{90}{98}-dfrac{91}{99}-dfrac{92}{100}}{dfrac{1}{45}+dfrac{1}{50}+dfrac{1}{55}+...+dfrac{1}{495}+dfrac{1}{500}}
Đọc tiếp
Tính giá trị biểu thức A , biết rằng A = M : N
Mà M = \(\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+\dfrac{4}{96}+...+\dfrac{97}{3}+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
N = \(\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)
Chứng minh rằng:
\(\dfrac{1}{6}\) < \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}\) < \(\dfrac{1}{4}\)
\(tính:\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{7}-\dfrac{1}{6}+\dfrac{1}{5}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{2}\)