limiting sum HELP (1 Viewer)

xXnukerrrXx · Mar 15, 2012

For the GP $(\sqrt5+\sqrt3)+(\sqrt5-\sqrt3)+...$ verify $S_\infty =T_1+\frac{1}{3}\sqrt3$

starryblue · Mar 15, 2012

what exercise, what question is that from textbk...i like did that qu, today.

Carrotsticks · Mar 15, 2012

Assuming this is a GP, the ratio can be found by putting the second term over the first

$\frac{\sqrt5 - \sqrt 3}{\sqrt5 + \sqrt3} = 4-\sqrt15$

So using limiting sum formula:

$S_\infty = \frac{\sqrt5 + \sqrt3}{1-(4-\sqrt{15})} = - \frac{\sqrt5 + \sqrt3}{3-\sqrt{15}}$

Rationalise the denominator and simplifying yields:

$\frac{3\sqrt5 + 4\sqrt3}{3} = \frac{3\sqrt5 + 3\sqrt3 + \sqrt3}{3} = \sqrt5 + \sqrt3 + \frac{\sqrt3}{3} = T_1 + \frac{\sqrt{3}}{3}$

limiting sum HELP (1 Viewer)

xXnukerrrXx

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starryblue

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Carrotsticks

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