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(x^k)  1 = (x  1)*(x^(k1) + x^(k2) + ... + x + 1)by General_Sax
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#1
Feb912, 09:25 PM

P: 450

(x^k)  1 = (x  1)*(x^(k1) + x^(k2) + ... + x + 1)
Where does this factorization come from? I need to know so I can use it in a proof. Thanks. 


#3
Feb912, 10:33 PM

P: 3,014

Use mathematical induction:
[tex] x^{k+1}  1 = (x^{k+1}  x^k) + (x^k  1) = (x  1) x^k +(x^k  1) [/tex] 


#4
Feb912, 11:09 PM

P: 450

(x^k)  1 = (x  1)*(x^(k1) + x^(k2) + ... + x + 1)
@Dickfore I'm confused as to the next step  yes I've been trying to work it out. Should I try to factor (x1) out of the expression? 


#5
Feb1012, 09:07 AM

P: 3,014




#6
Feb1012, 10:07 AM

HW Helper
P: 1,965

Multiply your last bracket by x on the next line multiply that same last bracket by 1. Compare. Add. This is a very useful formula. That last bracket is the geometrical series for example, hence you calculate it equals (x^{k}  1)/(x  1). Heard anything like that before? 


#7
Feb1012, 12:55 PM

P: 450

Thanks for the help people. I think I've got it. Just used formula for geometric series and did some algebra  hope it's good enough.



#8
Feb1012, 03:50 PM

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For all real numbers x such that x<1, ##\sum_{k=0}^\infty x^k=\frac{1}{1x}.##This theorem is much harder to understand than the formula you're trying to prove, because it involves convergence. Its standard proof uses the formula you're trying to prove, which by the way holds for all real numbers x, not just the ones that satisfy x<1. 


#9
Feb1012, 08:21 PM

P: 450

Thanks for the additional effort/attention Fredrik, but I've already used the geometric "series"  perhaps it's more accurate to use the term "sum"  in a proof.
http://en.wikipedia.org/wiki/Geometric_series#Formula that's the one I used. Just split it up w/ some algebra. It's for a CMPUT course and we haven't much experience with proofs so I don't think they expect much rigour. 


#10
Feb1112, 05:26 AM

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So basically you took the formula ##1+x+\cdots+x^{k1}=\frac{1x^{k}}{1x}## (which doesn't hold for x=1 by the way) and just multiplied it by x1? If I was your teacher, I wouldn't accept that as an answer. (The formula you found is too similar to what you're supposed to prove). Why don't you just do the multiplication on your righthand side to see what you get? Do you know how to evaluate a(b+c)? How about (a+b)(b+c)? How about (a+b)(b+c+d)?



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