(1/(2+(√2))) +(1/(3(√2)+2(√3))) +(1/(4(√3)+3(√4)))+...+(1/(100(√(99))+99(√(100))))


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Question Number 116226 by bobhans last updated on 02/Oct/20

$\frac{1}{2 + \sqrt{2}} + \frac{1}{3 \sqrt{2} + 2 \sqrt{3}} + \frac{1}{4 \sqrt{3} + 3 \sqrt{4}} + . . . + \frac{1}{100 \sqrt{99} + 99 \sqrt{100}}$
Commented by bemath last updated on 02/Oct/20

$0.9$
	Answered by john santu last updated on 02/Oct/20

	$\frac{1}{2 + \sqrt{2}} \times \frac{2 - \sqrt{2}}{2 - \sqrt{2}} = \frac{2 - \sqrt{2}}{2} = 1 - \frac{\sqrt{2}}{2}$ $\frac{1}{3 \sqrt{2} + 2 \sqrt{3}} \times \frac{3 \sqrt{2} - 2 \sqrt{3}}{3 \sqrt{2} - 2 \sqrt{3}} = \frac{3 \sqrt{2} - 2 \sqrt{3}}{6} = \frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{3}$ $\frac{1}{4 \sqrt{3} + 3 \sqrt{4}} \times \frac{4 \sqrt{3} - 3 \sqrt{4}}{4 \sqrt{3} - 3 \sqrt{4}} = \frac{4 \sqrt{3} - 3 \sqrt{4}}{12} = \frac{\sqrt{3}}{3} - \frac{\sqrt{4}}{4}$ $t h e n$ $= 1 - \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{3} + \frac{\sqrt{3}}{3} - \frac{\sqrt{4}}{4} + . . . + \frac{\sqrt{99}}{99} - \frac{\sqrt{100}}{100}$ $(t e l e s c o p i n g s e r i e s)$ $= 1 - \frac{\sqrt{100}}{100} = 1 - \frac{1}{10} = \frac{9}{10} = 0.9$
	Answered by Dwaipayan Shikari last updated on 02/Oct/20

	$\frac{1}{2 + \sqrt{2}} = \frac{2 - \sqrt{2}}{2} = 1 - \frac{1}{\sqrt{2}}$ $\frac{1}{3 \sqrt{2} + 2 \sqrt{3}} = \frac{3 \sqrt{2} - 2 \sqrt{3}}{6} = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}}$ $So$ $\frac{1}{2 + \sqrt{2}} + \frac{1}{3 \sqrt{2} + 2 \sqrt{3}} + . . . = 1 - \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}} + . . . . . - \frac{1}{\sqrt{100}}$ $= 1 - \frac{1}{10} = \frac{9}{10}$
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