Question 995916
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The &nbsp;<U>remainder theorem</U>&nbsp; says &nbsp;(see the lesson <A HREF=http://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/Divisibility-of-polynomial-f%28x%29-by-binomial-x-a.lesson>Divisibility of polynomial f(x) by binomial x-a</A>&nbsp; in this site)&nbsp; that the remainder of division the polynomial &nbsp;{{{f(x)}}}&nbsp; by 

the binomial &nbsp;{{{x-a}}}&nbsp; is equal to the value &nbsp;{{{f(a)}}}&nbsp; of the polynomial.


By applying this theorem to the given polynomial f(x) = {{{x^100 + 2x^99 +3x^98 +... 100x}}}, it means that the remainder from division the polynomial by &nbsp;{{{x-1}}}&nbsp; is equal 

to the value &nbsp;{{{f(1)}}}&nbsp; of the polynomial.


In turn, this value is 


f(1) = {{{1^100 + 2*1^99 +3*1^98 + . . . + 100*1}}} = 1 + 2 + 3 + . . . + 100.


It is the sum of first &nbsp;100 &nbsp;natural numbers = sum of &nbsp;100&nbsp; terms of the arithmetic progression with the first term &nbsp;1&nbsp; and the common difference &nbsp;1.


This sum is equal to &nbsp;{{{(100*101)/2}}} = 50*101 = 5050.


So, &nbsp;the required remainder is 5050.