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^p$$b^q$$c^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+$x^2$) = $\frac{1-x^3}{(1-x)}$, 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.