Question

Chứng minh rằng 1\2 mu 2+1\4 mu 2+1\6 mu 2+.....+1\4010 mu 2<1\2

Answer 1

giup minh lam nhanh nhanh len minh can gap ai la dung minh se k cho

Answer 2

\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{4010^2}\)

= \(\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2005^2}\right)\)

< \(\frac{1}{2^2}.\left(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2004.2005}\right)\)

\(=\frac{1}{2^2}.\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2004}-\frac{1}{2005}\right)\)

= \(\frac{1}{2^2}.\left(2-\frac{1}{2005}\right)=\frac{1}{2}-\frac{1}{4\left(2005\right)}< \frac{1}{2}\)

Vậy \(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{4010^2}< \frac{1}{2}\)