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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 ; B= \(1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-....-\frac{1}{2004^2}\)>\(\frac{1}{2004}\)
Chứng minh
B = \(1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-...-\frac{1}{2004^2}\) > \(\frac{1}{2004}\)
Cho \(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\) và \(x^2+y^2=1\) Chứng minh rằng: \(\frac{x^{2004}}{a^{1002}}+\frac{y^{2004}}{b^{1002}}=\frac{2}{\left(a+b\right)^{102}}\)
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 < \(\frac{1}{2}\)
1. Tính : P =\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{5}}{\frac{2004}{1}+\frac{2003}{2}+\frac{2002}{3}+...+\frac{1}{2004}}\)
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 \(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\)và \(x^2+y^2=1\)
Chứng minh : \(\frac{x^{2004}}{a^{1002}}+\frac{y^{2004}}{b^{1002}}=\frac{2}{\left(a+b\right)^{1002}}\)