# Solution to previous Problem of the Week

What is the remainder when f(x)=999x999+998x998+997x997+. . .+2x2+x is divided by x-1?

This is a finite polynomial of degree 999. For us to solve this problem we need to know that f(a) is the remainder when f(x) is divided by x-a. That is the famous remainder theorem.

The remainder when f(x)=999x999+998x998+997x997+. . .+2x2+x  is divided by x-1 is also f(1).

f(1)= 999(1)999+998(1)998+997(1)997+. . .+2(1)2+(1)

f(1)= 999+998+997+. . . +2+1

Using the formula for the sum of arithmetic series we have,

$f(1)=\displaystyle\frac{n(a_1+a_n)}{2}$ where a1 and an are the first and last term respectively, n is the number of terms.

f(1)=999(1+999)/2

f(1)=499500

Here is the link to that page

### 1 Response

1. Lester Salman M. Dahman says:

prove the following.in THEORY OF EQUATION
1. If R; +,∙ ›is a ring , then for any a Є R,0 ∙a = a ∙0=0, where 0 is the additive identity.
2. Let R be an abelian ring, then Øc : R[x]→R such that Øc(p(x)) = p(c) is a ring homomorphism

Can you help me to answer this question.Thank you