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Question Number 213047 by hardmath last updated on 29/Oct/24

$$\mathbf{n}=15$$$$\mathrm{find}:\frac{{4\mathrm{n}}^2+4\mathrm{n}+120}{\sqrt{{\mathrm{n}}^4+{2\mathrm{n}}^3+{\mathrm{n}}^2}}=?$$

Answered by mehdee7396 last updated on 29/Oct/24

$$\frac{{(2n+1)}^2+119}{n(n+1)}=4.5$$

Answered by MrGaster last updated on 29/Oct/24

$$\frac{4\cdot {15}^2+4\cdot 15+120}{\sqrt{{15}^4+2\cdot {15}^3+{15}^2}}$$$$\frac{4\cdot 225+60+120}{\sqrt{50625+6750+225}}$$$$\frac{900+60+120}{\sqrt{57600}}$$$$\frac{1080}{240}$$$$=4.5$$

Commented by hardmath last updated on 03/Nov/24

$$\mathrm{thank}\mathrm{you}\mathrm{dear}\mathrm{professors}$$

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