Challenge:

$f(x)=((x+3)^2)^{50}-4x+3=(x+3)^{100}-4x+3$

$(r_i+3)^{100}-4r_i+3=0 (i=1, 2, ...,100)$

$(r_i+3)^{100}=4r_i-3$

Therefore, $(r_1+3)^{100}+(r_2+3)^{100}+...+(r_{100}+3)^{100}=(4r_1-3)+(4r_2-3)+...+(4r_{100}-3)$

$=4(r_1+r_2+...+r_{100})-300$

$=4*(-300)-300=-1500$

(By Vieta's formulas, $-(r_1+r_2+...+r_{100})$ equals the coefficient of $x^{99}$ of $f(x)$, which is $nCr(100, 1)*3^1=300$ by the binomial theorem.

Therefore, $r_1+r_2+...+r_{100}=-300$.)