Question

Cu cat este egal A ???/ Ajutor este urgent ! 99 puncte

Answer 1

$Termenii~de~sub~radical~sunt~de~forma~ \frac{(2k-1)^2(2k+1)-2}{(2k-1)(2k+1)} = \\ \\ = \frac{(2k-1)^2(2k+1)}{(2k-1)(2k+1)} - \frac{2}{(2k-1)(2k+1)}= 2k-1- \frac{(2k+1)-(2k-1)}{(2k-1)(2k+1)}= \\ \\ =2k-1- \frac{1}{2k-1}+ \frac{1}{2k+1} . \\ \\ Deci~A= \sqrt{(1-1+ \frac{1}{3})+(3- \frac{1}{3}+ \frac{1}{5})+...+( 17- \frac{1}{17}+ \frac{1}{19}) }=\\ \\=\sqrt{(1+3+...+17)+(-1+ \frac{1}{3}- \frac{1}{3}+ \frac{1}{5}+...- \frac{1}{17}+ \frac{1}{19}) } = \\ \\=\sqrt{9^2-1+ \frac{1}{19} } =$

$=\sqrt{80+ \frac{1}{19} }= \\ \\ = \frac{39 \sqrt{19}}{19}.$