harder series and sequences (1 Viewer)

fannymasta · Aug 5, 2009

I need help with this question !
http://img512.imageshack.us/img512/5732/img009qjr.jpg

can someone please teach me how to find the answer :O or show me

THANKS !

Timothy.Siu · Aug 5, 2009

if u look, you can see that its a geometric series going vertically down

Mountain.Dew · Aug 5, 2009

Actually, it's gonna be a series of geometric series.

The answer that I've got is (3x - 1)/[(x-1)^2], assuming that |x| < 1.

kurt.physics · Aug 5, 2009

Mountain.Dew said:
Actually, it's gonna be a series of geometric series.

The answer that I've got is (3x - 1)/[(x-1)^2], assuming that |x| < 1.

How did you get a numerator of 3x - 1, i got x + 1?

kurt.physics · Aug 6, 2009

My solution:

$1 + 3x + 5x^2 + 7x^3 + 9x^4 + . . .$

$1 + (x + 2x) + (x^2 + 2x^2 + 2x^2) + (x^3 + 2x^3 + 2x^3 + 2x^3) + (x^4 + 2x^4 + 2x^4 + 2x^2 + 2x^2) + . . .$

$(1 + x + x^2 + x^3 + x^4 + . . . ) + (2x + 2x^2 + 2x^3 + 2x^4 + . . . ) + (2x^2 + 2x^3 + 2x^4 + 2x^5 . . . ) + (2x^3 + 2x^4 + 2x^5 + . . .) + (2x^4 + 2x^5 + 2x^6 + . . . ) + . . .$

$S_{∞} = \frac{a}{1-r}\,\,\,|r|<1$

$\therefore \frac{1}{1 - x} + \frac{2x}{1 - x} + \frac{2x^2}{1 - x} + \frac{2x^3}{1 - x} + . . .$

$= \frac{1}{1-x} + \frac{2x + 2x^2 + 2x^3 + 2x^4 + . . . }{1 - x}$

$= \frac{1}{1-x} + \left(2x + 2x^2 + 2x^3 + 2x^4 + . . .\right) \cdot \frac{1}{1-x}$

$= \frac{1}{1-x} + \frac{2x}{1-x} \cdot \frac{1}{1-x}$

$= \frac{1}{1-x} + \frac{2x}{(1 - x)^2}$

$= \frac{1-x}{(1-x)^2} + \frac{2x}{(1 - x)^2}$

$= \frac{1 - x + 2x}{(1 - x)^2}$

$= \boxed{\frac{x + 1}{(1 - x)^2}}$

Mountain.Dew · Aug 6, 2009

Yes, my bad, you are correct. it is x+1 rather than 3x - 1

harder series and sequences (1 Viewer)

fannymasta

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Timothy.Siu

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Mountain.Dew

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kurt.physics

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kurt.physics

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Mountain.Dew

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