Question Number 137588 by bramlexs22 last updated on 04/Apr/21 | ||
$${For}\:{a}\:{positive}\:{number}\:{n}\:,\:{let} \\ $$$${f}\left({n}\right)\:{be}\:{the}\:{value}\:{of}\: \\ $$$${f}\left({n}\right)=\frac{\mathrm{4}{n}+\sqrt{\mathrm{4}{n}^{\mathrm{2}} −\mathrm{1}}}{\:\sqrt{\mathrm{2}{n}+\mathrm{1}}\:+\sqrt{\mathrm{2}{n}−\mathrm{1}}} \\ $$$${calculate}\:{f}\left(\mathrm{1}\right)+{f}\left(\mathrm{2}\right)+{f}\left(\mathrm{3}\right)+...+{f}\left(\mathrm{40}\right). \\ $$ | ||
Answered by bemath last updated on 04/Apr/21 | ||
$$\Leftrightarrow\:{f}\left({n}\right)=\frac{\left(\sqrt{\mathrm{2}{n}+\mathrm{1}}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{2}{n}−\mathrm{1}}\right)^{\mathrm{2}} +\sqrt{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}−\mathrm{1}\right)}}{\:\sqrt{\mathrm{2}{n}+\mathrm{1}}\:+\sqrt{\mathrm{2}{n}−\mathrm{1}}} \\ $$$${f}\left({n}\right)=\:\frac{\left(\sqrt{\mathrm{2}{n}+\mathrm{1}}\:\right)^{\mathrm{3}} −\left(\sqrt{\mathrm{2}{n}−\mathrm{1}}\:\right)^{\mathrm{3}} }{\mathrm{2}} \\ $$$${f}\left(\mathrm{1}\right)+{f}\left(\mathrm{2}\right)+{f}\left(\mathrm{3}\right)+...+{f}\left(\mathrm{40}\right) \\ $$$$=\:\frac{\sqrt{\mathrm{3}^{\mathrm{3}} }−\sqrt{\mathrm{1}^{\mathrm{3}} }}{\mathrm{2}}\:+\:\frac{\sqrt{\mathrm{5}^{\mathrm{3}} }−\sqrt{\mathrm{3}^{\mathrm{3}} }}{\mathrm{2}}\:+\:\frac{\sqrt{\mathrm{7}^{\mathrm{3}} }−\sqrt{\mathrm{5}^{\mathrm{3}} }}{\mathrm{2}}\:+...+\:\frac{\sqrt{\mathrm{81}^{\mathrm{3}} }−\sqrt{\mathrm{79}^{\mathrm{3}} }}{\mathrm{2}} \\ $$$$\left[\:{telescopy}\:{series}\:\right] \\ $$$${f}\left(\mathrm{1}\right)+{f}\left(\mathrm{2}\right)+{f}\left(\mathrm{3}\right)+...+{f}\left(\mathrm{40}\right) \\ $$$$=\:\frac{\sqrt{\mathrm{81}^{\mathrm{3}} }−\sqrt{\mathrm{1}^{\mathrm{3}} }}{\mathrm{2}}\:=\:\frac{\mathrm{9}^{\mathrm{3}} −\mathrm{1}}{\mathrm{2}}\:=\:\frac{\mathrm{728}}{\mathrm{2}}\:=\:\mathrm{364}\: \\ $$ | ||