FMM18 - Revision history

Admin at 22:22, 24 January 2020

2020-01-24T22:22:07Z

← Older revision		Revision as of 22:22, 24 January 2020
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	</math>		</math>

	We also note that <math>a = b + 1<.math>, so <math>a - b = 1</math> and the whole thing reduces to		We also note that <math>a = b + 1</math>, so <math>a - b = 1</math> and the whole thing reduces to

	<math>		<math>

2020-01-23T02:46:16Z

2020-01-23T02:41:09Z

← Older revision		Revision as of 02:41, 23 January 2020
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	\|~~answer~~=		\|solution=
	There are 1977 terms to add, and surely it would not be a Friday Morning Math Problem if it were so boring as to tediously add so many terms.		There are 1977 terms to add, and surely it would not be a Friday Morning Math Problem if it were so boring as to tediously add so many terms.

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	<math>		<math>
	\dfrac{1}{\sqrt{a} + \sqrt{b}} = \dfrac{(\sqrt{a} - \sqrt{b})}{(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b})}		\dfrac{1}{\sqrt{a} + \sqrt{b}}
			= \dfrac{(\sqrt{a} - \sqrt{b})}{(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b})}
			= \dfrac{\sqrt{a} - \sqrt{b}}{\sqrt{a}^2 - \sqrt{b}^2}
			= \dfrac{\sqrt{a} - \sqrt{b}}{a - b}
	</math>		</math>

			We also note that <math>a = b + 1<.math>, so <math>a - b = 1</math> and the whole thing reduces to

			<math>
			\sqrt{1978} - 1
			</math>


	}}		}}

2020-01-23T02:23:56Z

← Older revision		Revision as of 02:23, 23 January 2020
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	\|answer=		\|answer=
			There are 1977 terms to add, and surely it would not be a Friday Morning Math Problem if it were so boring as to tediously add so many terms.

			You'll also notice the radicals are improper. If we try getting rid of the radicals on the bottom of each term, we get something interesting:

			<math>
			\dfrac{1}{\sqrt{a} + \sqrt{b}} = \dfrac{(\sqrt{a} - \sqrt{b})}{(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b})}
			</math>

	}}		}}

2020-01-23T02:15:35Z

← Older revision		Revision as of 02:15, 23 January 2020
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	{{FMM		{{FMM
	\|title=~~Petersburg Paradox Revisited~~		\|title=Radical Sums
	\|problem=		\|problem=
			Evaluate the sum:

			<math>
			\dfrac{1}{1 + \sqrt{1}}
			+ \dfrac{1}{\sqrt{2} + \sqrt{3}}
			+ \dfrac{1}{\sqrt{3} + \sqrt{4}}
			+ \dots
			+ \dfrac{1}{\sqrt{1977} + \sqrt{1978}}
			</math>


	\|answer=		\|answer=

	}}		}}

2019-12-04T10:34:59Z

Created page with "{{FMM |title=Petersburg Paradox Revisited |problem= |answer= }}"

New page

{{FMM
|title=Petersburg Paradox Revisited
|problem=

|answer=

}}