Question:
Find the coefficient of $x^{11}$ in the expansion of:$$(1+x^2)^4(1+x^3)^7(1+x^4)^{12}$$
The traditional way of doing this, as far as I know, is to first find the coefficient of each term that has $x^{11}$, and then sum it. However, with three individual terms in multiplication, this would be a very tedious job. Is there a shorter way to approach such problems?
On
You can find the coefficient of whichever term you wish by using a Taylor expansion. Indeed, the binomial series is just the Maclaurin series for $(1 + x)^n$ (i.e. the Taylor expansion about $x = 0$).
Doing so you'll find that the coefficient of $x^{11}$ is $1113$.
On
To calculate the coefficient by hand is somewhat cumbersome. Here is a notation which helps to make it better manageable. It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series.
We obtain by successively applying the binomial theorem \begin{align*} [x^{11}]&(1+x^2)^4(1+x^3)^7(1+x^4)^{12}\\ &=[x^{11}]\sum_{j=0}^{12}\binom{12}{j}x^{4j}(1+x^2)^4(1+x^3)^7\tag{1}\\ &=\sum_{j=0}^{2}\binom{12}{j}[x^{11-4j}](1+x^2)^4(1+x^3)^7\tag{2}\\ &=\sum_{j=0}^{2}\binom{12}{j}[x^{11-4j}]\sum_{k=0}^7\binom{7}{k}x^{3k}(1+x^2)^4\tag{3}\\ &=\sum_{j=0}^{2}\sum_{k=0}^3\binom{12}{j}\binom{7}{k}[x^{11-4j-3k}](1+x^2)^4\qquad\qquad\quad 4j+3k\leq 11\tag{4}\\ &=\sum_{j=0}^{2}\sum_{k=0}^3\binom{12}{j}\binom{7}{k}[x^{11-4j-3k}] \sum_{l=0}^{4}\binom{4}{l}x^{2l}\qquad\qquad 4j+3k\leq 11\tag{5}\\ &=\sum_{j=0}^{2}\sum_{k=0}^3\sum_{l=0}^{4} \binom{12}{j}\binom{7}{k}\binom{4}{l}\qquad\qquad\qquad\qquad 4j+3k+2l= 11\tag{6}\\ \end{align*}
Comment:
In (1) we start expanding the binomial with the greatest power $x^4$ which keeps the number of summands small as we will see in the next step.
In (2) we use the linearity of the coefficient of operator and apply the rule $[x^{m-n}]=[x^m]x^n$. Note that since the power of $x$ is non negative, the upper limit of the index $j$ is now $2$.
In (3) we expand the binomial $(1+x^3)^7$.
In (4) we do a similar job as in (2) and keep in mind, that the power of $x$ is non-negative by stating $4j+3k\leq 11$.
In (5) we expand the last binomial
In (6) we apply the coefficient of operator the last time and obtain a summation formula for the coefficient of $x^{11}$.
The final step is to do some bookkeeping and look according to (6) for all triples $(j,k,l)$ with
\begin{align*} \begin{matrix} 0\leq j \leq 2\\ 0\leq k \leq 3\\ 0\leq l \leq 4\\ \end{matrix} \qquad\quad\text{and}\quad\qquad 4j+3k+2l= 11 \end{align*}
We find \begin{array}{rrrrr} j&k&l&\binom{12}{j}\binom{7}{k}\binom{4}{l}&\sum\\ \hline 2&1&0&462&462\\ 1&1&2&504&966\\ 0&1&4&7&973\\ 0&3&1&140&\color{blue}{1113}\\ \end{array}
resulting finally in
\begin{align*} [x^{11}](1+x^2)^4(1+x^3)^7(1+x^4)^{12}=1113 \end{align*}
in accordance with the result of Wolfram Alpha.
Based on your equation, I feel like the only way you can get $x^{11}$ is $(x^2)^4 * x^3 *constant$ or $(x^2)^2 * x^3 * x^4$ or $x^2 * (x^3)^2 * constant$ or $constant * x^3 * (x^4)^2$. Then you can use formula for binomial expansion to calculate the coefficients for each term. Hopefully it helps.