Help: Expansion

Hi there today I stumbled upon a question solution which uses the following expansion.

11x=1+x+x2+x3+...\dfrac{1}{1-x} = 1 + x + x^2 + x^3 + ... provided x<1|x| < 1

I guess I am forgetting a really simple step for this so I would like to know how this relation is obtained.

#Algebra

Note by Sanjay Swain
9 months, 2 weeks ago

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Comments

This is the formula of a geometric series of first term, a=1a=1 and common ratio r=xr=x.

The formula says:

If first term =a=a and common ratio =r,r<1=r, |r|<1,

the sum of geometric series is:

a1r\dfrac{a}{1-r}

Substitute values, we get:

a1r=11x=1+x+x2+x3\dfrac{a}{1-r}=\dfrac{1}{1-x}=1+x+x^2+x^3 \cdots

Vinayak Srivastava - 9 months, 2 weeks ago

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Oh thanks I guess that's why I found this expression familiar. Guess I need to revisit geometric series. :D

Sanjay Swain - 9 months, 2 weeks ago
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