5. The coefficient of $x^{150}$ in the expansion of $\left(1+x+x^{2}+\cdots+x^{100}\right)^{3}$, after combining like terms, is $\qquad$
5.7651 .

By the polynomial multiplication rule, the problem can be transformed into finding the number of natural number solutions of the equation
$$
s+t+r=150
$$

that do not exceed 100.
Obviously, the number of natural number solutions of equation (1) is $\mathrm{C}_{152}^{2}$.
Next, we find the number of natural number solutions of equation (1) that exceed 100.
Since their sum is 150, there can only be one number greater than 100.
Assume without loss of generality that $s>100$, and transform equation (1) into
$$
(s-101)+t+r=49 \text {. }
$$

Let $s^{\prime}=s-101$.
Then the number of natural number solutions of the equation $s^{\prime}+t+r=49$ is $\mathrm{C}_{51}^{2}$.
Therefore, the coefficient of $x^{150}$ is $\mathrm{C}_{52}^{2}-\mathrm{C}_{3}^{1} \mathrm{C}_{51}^{2}=7651$.