Question

Cho biểu thức \(D=\frac{\left(2!\right)^2}{1^2}+\frac{\left(2!\right)^2}{3^2}+\frac{\left(2!\right)^2}{5^2}+\frac{\left(2!\right)^2}{7^2}+.....+\frac{\left(2!\right)^2}{2015^2}\)

Soanhs D với 6

Nhanh nhanh giúp mình nhé. Thanh you! 

Answer 1

Ta có: \(D=2\left(\frac{2}{1^2}+\frac{2}{3^2}+...+\frac{2}{2015^2}\right)< 2\left(2+\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2013.2015}\right)\)

\(=2\left(2+1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2013}-\frac{1}{2015}\right)=2\left(3-\frac{1}{2015}\right)=6-\frac{2}{2015}\)

Vậy D < 6.

Answer 2

\(D=\frac{\left(2!\right)^2}{1^2}+\frac{\left(2!\right)^2}{3^2}+\frac{\left(2!\right)^2}{5^2}+\frac{\left(2!\right)^2}{7^2}+...+\frac{\left(2!\right)^2}{2015^2}\)

=>\(D=\frac{\left(1.2\right)^2}{1^2}+\frac{\left(1.2\right)^2}{3^2}+\frac{\left(1.2\right)^2}{5^2}+\frac{\left(1.2\right)^2}{7^2}+...+\frac{\left(1.2\right)^2}{2015^2}\)

=>\(D=\frac{2^2}{1^2}+\frac{2^2}{3^2}+\frac{2^2}{5^2}+\frac{2^2}{7^2}+...+\frac{2^2}{2015^2}\)

=>\(D=2\left(\frac{2}{1^2}+\frac{2}{3^2}+\frac{2}{5^2}+\frac{2}{7^2}+...+\frac{2}{2015^2}\right)\)

Ta có: \(\frac{2}{1^2}+\frac{2}{3^2}+\frac{2}{5^2}+\frac{2}{7^2}+...+\frac{2}{2015^2}< 2+\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\)

=>\(D=2\left(\frac{2}{1^2}+\frac{2}{3^2}+\frac{2}{5^2}+\frac{2}{7^2}+...+\frac{2}{2015^2}\right)< 2\left(2+\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)\)

Mà \(2\left(2+\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)\)\(=2\left(2+\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)

\(=2\left(2+1-\frac{1}{2015}\right)=2\left(3-\frac{1}{2015}\right)=6-\frac{6}{2016}< 6\)

=>\(D< 2\left(2+\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)< 6\)

=>D<6