Các câu hỏi tương tự
Cho S=\(\frac{1}{5^2}-\frac{2}{5^3}+\frac{3}{5^4}-\frac{4}{5^5}+...+\frac{99}{5^{100}}-\frac{100}{5^{101}}\)
Chứng minh rằng \(S< \frac{1}{36}\)
Tính:
\(\frac{1}{2}-\left(\frac{1}{3}+\frac{2}{3}\right)+\left(\frac{1}{4}+\frac{2}{4}+\frac{3}{4}\right)-\left(\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}\right)+...+\left(\frac{1}{100}+\frac{2}{100}+\frac{3}{100}+...+\frac{99}{100}\right)\)
Tính các tổng sau:
a) \(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}.\)
b) \(-\frac{4}{5}+\frac{4}{5^2}-\frac{4}{5^3}+...+\frac{4}{5^{200}}.\)
c)\(\frac{-1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{100}}-\frac{1}{3^{101}}\)
CMR:
a, \(100-\left(1+\frac{1}{2}+\frac{1}{3}+..+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+..+\frac{99}{100}\)
b, \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+..+\frac{1}{200}\right)=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
Giải nhanh giùm mình nhé!!!!!!!!!!!!!!
So sánh M và N, biết
\(M=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}\)và \(N=\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\)
1, Tính \(\frac{1}{2}-\left(\frac{1}{3}+\frac{2}{3}\right)+\left(\frac{1}{4}+\frac{2}{4}+\frac{3}{4}\right)-\left(\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}\right)+...+\left(\frac{1}{100}+\frac{2}{100}+\frac{3}{100}+...+\frac{99}{100}\right)\)2,Tính \(\left(1-\frac{1}{2^2}\right)x\left(1-\frac{1}{3^2}\right)x\left(1-\frac{1}{4^2}\right)x...x\left(1-\frac{1}{n^2}\right)\)
cho \(M=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\)
\(N=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\)
CMR: M < 1/10
Cho \(M=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.........\frac{99}{100}\)và \(N=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}.......\frac{100}{101}\)Tính tích M.N
Ai nhanh mik tick cho
Cho M =\(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}vaN=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\)
a) Tinh tich M.N
b) chung minh M<N
c) Chung minh M < \(\frac{1}{10}\)