Các câu hỏi tương tự
Chứng tỏ rằng :\(B=1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{3^2}-...-\frac{1}{2004^2}>\frac{1}{2004}\)
Chứng minh rằng \(B=1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-...-\frac{1}{2004^2}>\frac{1}{2004}\)
Chứng tỏ rằng ;S=\(\frac{1}{2^2}-\frac{1}{2^4}+...+\frac{1}{2^{2002}}-\frac{1}{2^{2004}}\)<0,2
chứng minh rằng :\(\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+.........+\frac{2004}{4^{2004}}
Chứng minh
B = \(1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-...-\frac{1}{2004^2}\) > \(\frac{1}{2004}\)
Bài 2 : Tính : B = \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2005}}{\frac{2004}{1}+\frac{2003}{2}+\frac{2002}{3}+...+\frac{1}{2004}}\)
Tính:
B=\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..+\frac{1}{2005}}{\frac{2004}{1}+\frac{2003}{2}+\frac{2002}{3}+..+\frac{1}{2004}}\)
cho B =\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\)chứng minh rằng B < \(\frac{1}{2}\)
Tính
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2015}}\)
\(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2005}}{\frac{2004}{1}+\frac{2003}{2}+\frac{2002}{3}+...+\frac{1}{2004}}\)