The multinomial theorem generalizes the binomial: (x_1+...+x_k)^n = sum [n!/(r_1!...r_k!)] * x_1^{r_1}...x_k^{r_k}, summed over all non-negative integers r_1,...,r_k with r_1+...+r_k = n.

The multinomial coefficient n!/(r_1!...r_k!) counts the number of permutations of n objects with r_i objects of type i. The number of terms in the expansion is C(n+k-1, k-1) by stars and bars.

For trinomials (k=3): (a+b+c)^n has C(n+2, 2) terms. The coefficient of a^pb^qc^r (where p+q+r=n) is n!/(p!*q!*r!).

Practical approach for JEE: Most problems involve finding a specific coefficient in (1+x+x^2)^n or similar. Method: identify all tuples (r_1,...,r_k) that give the desired power of x, compute the multinomial coefficient for each, and sum.

Alternative factorization: (1+x+x^2) = (1-x^3)/(1-x), so (1+x+x^2)^n = (1-x^3)^n*(1-x)^{-n}. This converts the multinomial into a product of two binomial series, often simplifying coefficient extraction.

The negative binomial series (1-x)^{-n} = sum_{r>=0} C(n+r-1,r)*x^r is essential for this approach. Combined with the finite expansion of (1-x^3)^n, coefficient extraction becomes a two-term convolution.