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Question Number 212219 by hardmath last updated on 06/Oct/24

$$\mathrm{Find}:$$$$\sqrt{21\cdot 22\cdot 23\cdot 24+1}=?$$

Answered by Rasheed.Sindhi last updated on 06/Oct/24

$$\sqrt{(22.5-1.5)(22.5-.5)(22.5+.5)(22.5+1.5)+1}$$$$\sqrt{(22.5^2-.5^2)(22.5^2-1.5^2)+1}$$$$\sqrt{506\cdot 504+1}$$$$\sqrt{(505+1)(505-1)+1}$$$$\sqrt{({505}^2-1^2)+1}$$$$\sqrt{{505}^2-1+1}$$$$\sqrt{{505}^2}$$$$505$$

Commented by racer last updated on 13/Oct/24

hh

Answered by Frix last updated on 06/Oct/24

$$x(x+1)(x+2)(x+3)+1={(x^2+3x+1)}^2$$$$x=21\Rightarrow \mathrm{answer}\mathrm{is}505$$

Answered by Rasheed.Sindhi last updated on 07/Oct/24

$$\sqrt{21\cdot 22\cdot 23\cdot 24+1}$$$$=\sqrt{\frac{42}{2}\cdot \frac{44}{2}\cdot \frac{46}{2}\cdot \frac{48}{2}+1}$$$$=\sqrt{\frac{42.44.46.48}{2^4}+1}$$$$=\sqrt{\frac{(45-3)(45-1)(45+1)(45+3)}{16}+1}$$$$=\sqrt{\frac{({45}^2-1^2)({45}^2-3^2)}{16}+1}$$$$=\sqrt{\frac{2024\cdot 2016}{16}+1}$$$$=\sqrt{\frac{(2020+4)(2020-4)}{16}+1}$$$$=\sqrt{\frac{{2020}^2-4^2}{4^2}+1}$$$$=\sqrt{\frac{{2020}^2}{4^2}-1+1}$$$$=\sqrt{\frac{2020}{4}}$$$$=505$$

Commented by hardmath last updated on 08/Oct/24

$$\mathrm{thankyou}\mathrm{dear}\mathrm{professors}$$

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