What is the remainder when #1 + x + x^2 + x^3 + ... + x^2006# is divided by #(x-1)# ?

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Answer 1 · 2016-12-04T11:21:56

#2007#

If #f(x)# is a polynomial, then the remainder when dividing #f(x)# by #(x-a)# is #f(a)#

In our example:

#f(x) = 1 + x + x^2 + x^3 + ... + x^2006#

So the remainder when dividing by #(x-1)# is:

#f(1) = 1 + 1 + 1 + 1 + ... + 1 = 2007#

Answer 2 · 2016-12-10T22:20:11

#2007#

#1+x+x^2+cdots+x^2006=(x^2007-1)/(x-1)=q(x)(x-1)+C#

or

#x^2007-1=q(x)(x-1)^2+C(x-1)# now deriving both sides

#2007 x^2006=2q(x)(x-1)+q'(x)(x-1)^2+C#

Making now #x = 1# we have

#2007=C#

so the answer is #2007#