2014/111) Show that the function below has same remainder when divided by $x(x+1)$ and $x(x+1)^2$

$f(x)=2008+2007x+2006x^2+\cdots+3x^{2005}+2x^{2006}+x^{2007}$

1st we provide the premise

this shall have same remainder when divided by $x(x+1)$ and $x(x+1)^2$

provided this shall have same remainder when divided by $(x+1)$ and $(x+1)^2$

when we divide by (x+1)^2 we shall have a linear polynomial say m(x+1) + c and when we divide by (x+1) it shall be d and both are same if m= 0 and m+c = d

so if we convert

f(x) as a polynomial of (x+1) the coefficient of x should be zero

of f(x-1) should have coefficient of x to be zero

now we provide the solution based on premise

$f(x-1)=2008+2007(x-1)+2006(x-1)^2+\cdots$
   $+3(x-1)^{2005}+2(x-1)^{2006}+(x-1)^{2007}$

should have coefficient of x to be zero

the coefficient of x

$= 2007 + 2006 * (-2) + 2005 * 3 \cdots + 2 * 2006 - 2007 = 0$

hence proved

I have solved the problem at http://mathhelpboards.com/challenge-questions-puzzles-28/division-polynomial-13760.html#post65389 where you can find some more different solutions