Question Number 218438 by Marzuk last updated on 10/Apr/25 $${An}\:{amazing}\:{thing}\:{i}\:{saw} \\ $$$${S}\:=\:\mathrm{1}\:+\:\mathrm{2}\:+\:\mathrm{3}\:+\:\mathrm{4}\:+\:\mathrm{5}\:+\:\mathrm{6}… \\ $$$$\:\:\:\:=\:\mathrm{1}\:+\:\mathrm{2}\left(\mathrm{2}/\mathrm{2}\:+\:\mathrm{3}/\mathrm{2}\:+\:\mathrm{4}/\mathrm{2}\:+\:\mathrm{5}/\mathrm{2}\:+\mathrm{6}/\mathrm{2}….\right) \\ $$$$\:\:\:\:=\:\mathrm{1}\:+\:\mathrm{2}\left(\mathrm{1}\:+\:\mathrm{3}/\mathrm{2}\:+\:\mathrm{2}\:+\:\mathrm{5}/\mathrm{2}\:+\:\mathrm{3}…\right) \\ $$$$\:\:\:\:=\:\mathrm{1}\:+\:\mathrm{2}\left(\mathrm{1}+\:\mathrm{2}\:+\:\mathrm{3}\:…\:+\:\mathrm{3}/\mathrm{2}\:+\:\mathrm{5}/\mathrm{2}…\right) \\ $$$$\:\:\:\:=\:\mathrm{1}\:+\:\mathrm{2}{S}\:+\:\mathrm{2}\underset{{n}=\:\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}{n}\:+\:\mathrm{1}}{\mathrm{2}} \\ $$$${or},{S}\:−\:\mathrm{2}{S}\:=\:\mathrm{1}\:+\:\underset{{n}=\mathrm{1}} {\overset{\infty}…
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Question Number 218428 by hardmath last updated on 09/Apr/25 Answered by A5T last updated on 09/Apr/25 $$\mathrm{Let}\:\mathrm{F}\:\in\:\mathrm{BC}\:\mathrm{such}\:\mathrm{that}\:\mathrm{AF}\bot\mathrm{BC} \\ $$$$\mathrm{AB}=\mathrm{AE}\Rightarrow\mathrm{BF}=\mathrm{FE}=\frac{\mathrm{7}+\mathrm{3}}{\mathrm{2}}=\mathrm{5} \\ $$$$\mathrm{BF}×\mathrm{BC}=\mathrm{BA}^{\mathrm{2}} \Rightarrow\mathrm{5}\left(\mathrm{10}+\mathrm{x}\right)=\mathrm{BA}^{\mathrm{2}} …\left(\mathrm{i}\right) \\ $$$$\mathrm{AB}^{\mathrm{2}}…
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Question Number 218399 by Nicholas666 last updated on 09/Apr/25 Answered by MrGaster last updated on 10/Apr/25 $$\begin{array}{|c|}{{f}\left({n}\right)=\frac{{n}^{\mathrm{2}} }{\mathrm{2}}}\\\hline\end{array}\mathrm{when}\:{n}\:\mathrm{is}\:\mathrm{even} \\ $$$$\cancel{{f}\left({n}\right)=\frac{{n}^{\mathrm{2}} }{\mathrm{2}}} \\ $$$${f}\left({n}\right)=\frac{{n}^{\mathrm{2}} −\tau\left({n}\right)}{\mathrm{2}} \\…
Question Number 218393 by Nicholas666 last updated on 09/Apr/25 $$ \\ $$$$\:{let}\:{ABC}\:{be}\:{a}\:{triangle}\:{with}\:{incenter}\:{I}. \\ $$$$\:{prove}\:{that}\:{Ia}\:.\:{Ib}\:.\:{Ic}\:\:\leqslant\:\frac{{abc}}{\mathrm{8}}\:\: \\ $$$$ \\ $$ Answered by mr W last updated on…
Question Number 218427 by mathocean1 last updated on 09/Apr/25 $$ \\ $$$${give}\:{a}\:{recurrence}\:{relation}\:{for}\:{I}_{{n}} . \\ $$$${I}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}} }{dx},\:\forall{n}\:\in\:\mathbb{N}. \\ $$ Answered by mehdee7396…
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