$(1+x+x^2+\cdots+x^{100})^3$展开式中$x^{150}$前的系数为_____

解答:$(1+x+x^2+\cdots+x^{100})^3=(1-x^{101})^3\sum\limits_{k=0}^{+\infty}C_{k+2}^2x^k$
$=(1-3x^{101}+3x^{202}-x^{303})\sum\limits_{k=0}^{+\infty}C_{k+2}^2x^k$
所以$x^{150}$前的系数为$C_{152}^2-3C_{51}^2$