Animated Solution for Mathematics - Binomial Theorem: In the expansion of $(1+x)(1-x^2)(1+\frac{3}{x}+\frac{3}{x^2}+\frac{1}{x^3}{)}^5,x\ne 0$, the sum of the coefficient of $x^3$ and $x^{-13}$ is equal to _______.

Imagine you are standing before a massive, intimidating algebraic expression:

It looks like a trap, doesn't it? A classic JEE Advanced problem designed to make you panic and start calculating blindly. But take a deep breath; in the world of competitive exams, complexity is often just a mask for elegance.

Our first step is to hunt for patterns. Look at that third bracket: $\left(1+\frac{3}{x}+\frac{3}{x^2}+\frac{1}{x^3}\right)$.

If you have spent enough time with the Binomial Theorem, you will recognize this as the expansion of ${\left(1+\frac{1}{x}\right)}^3$. It follows the classic structure:

By identifying this, we have already tamed the beast. The expression now becomes:

Now, let's look at the base ${\left(1+\frac{1}{x}\right)}^{15}$. We can write this as ${\left(\frac{x+1}{x}\right)}^{15}$, which is $\frac{(1+x{)}^{15}}{x^{15}}$.

Next, consider the first two terms: $(1+x)(1-x^2)$. We know that $(1-x^2)$ is a difference of squares, $(1-x)(1+x)$.

So, our expression is now:

Gathering all the $(1+x)$ terms, we have $(1-x)(1+x{)}^{17}\times x^{-15}$. We have transformed a terrifying expression into a clean, manageable form:

The problem asks for the sum of the coefficients of $x^3$ and $x^{-13}$. Because we have an $x^{15}$ in the denominator, we cannot simply look for $x^3$ and $x^{-13}$ in the numerator.

We must shift our target: To get an overall power of $x^3$, the numerator must provide $x^{15+3}=x^{18}$. To get an overall power of $x^{-13}$, the numerator must provide $x^{15-13}=x^2$.

This is the "aha!" moment that separates the top rankers from the rest.

Now, we focus on the numerator: $(1-x)(1+x{)}^{17}=(1+x{)}^{17}-x(1+x{)}^{17}$. We need the coefficient of $x^{18}$ and $x^2$ in this expansion.

For $x^{18}$: The first part $(1+x{)}^{17}$ has a maximum power of $17$, so its coefficient for $x^{18}$ is $0$. The second part $-x(1+x{)}^{17}$ is $-1\times$ (coefficient of $x^{17}$ in $(1+x{)}^{17}$), which is $-1\times {}^{17}{C}_{17}=-1$. Thus, the coefficient of $x^3$ is $-1$.

For $x^2$: The first part $(1+x{)}^{17}$ gives ${}^{17}{C}_2=\frac{17\times 16}{2}=136$. The second part $-x(1+x{)}^{17}$ gives $-1\times$ (coefficient of $x^1$ in $(1+x{)}^{17}$), which is $-1\times {}^{17}{C}_1=-17$. Thus, the coefficient of $x^{-13}$ is $136-17=119$.

Final Calculation: The final sum is $119+(-1)=118$. We have conquered the monster! The final answer is 118.