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Question Number 69296 by ozodbek last updated on 22/Sep/19

Commented by Prithwish sen last updated on 22/Sep/19

$$\mathrm{can}\mathrm{you}\mathrm{post}\mathrm{the}\mathrm{answer}\mathrm{please}?$$

Commented by MJS last updated on 22/Sep/19

$$\mathrm{approximation}\mathrm{gives}8$$

Commented by Prithwish sen last updated on 22/Sep/19

$$8=3!+2!=\sqrt{{(3!+2!)}^2}=\sqrt{2!(2!+2.3!+3.3!)}$$$$=\sqrt{2!\{2!+2.3!+3!(4-1)\}}=\sqrt{2!\{2!+3!+4!)}$$$$=\sqrt{2!\{2!+\sqrt{{(3!+4!)}^2}}=\sqrt{2{!}^2+2!\sqrt{3!(3!+2.4!+4.4!)}}$$$$=\sqrt{2{!}^2+2!\sqrt{3!\{3!+2.4!+4!(5-1)\}}}$$$$=\sqrt{2{!}^2+2!\sqrt{3!\{3!+(4!+5!)}}$$$$=\sqrt{2{!}^2+2!\sqrt{3{!}^2+3!\sqrt{{(4!+5!)}^2}}}......$$$$=\sqrt{2{!}^2+2!\sqrt{3{!}^2+3!\sqrt{4{!}^2+4!\sqrt{.......}}}}$$$$\mathbf{Thanks}\mathbf{Sir}\mathbf{MJS}\mathbf{your}\mathbf{approximation}\mathbf{help}\mathbf{me}$$$$\mathbf{a}\mathbf{lot}.\mathbf{I}\mathbf{appreciate}\mathbf{your}\mathbf{cooperation}.\mathbf{Again}$$$$\mathbf{thanks}\mathbf{a}\mathbf{lot}.$$

Commented by MJS last updated on 22/Sep/19

$$\mathrm{great}!\mathrm{And}\mathrm{as}\mathrm{always},{\mathrm{you}}^{\prime }\mathrm{re}\mathrm{welcome}$$

Commented by ozodbek last updated on 23/Sep/19

$$\mathrm{thank}\mathrm{you}$$

Commented by Prithwish sen last updated on 23/Sep/19

$$\mathrm{you}\mathrm{are}\mathrm{welcome}.$$

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