Animated Solution for Mathematics - Binomial Theorem: If the coefficients of $x$ and $x^2$ in $(1+x{)}^p(1-x{)}^q$ are 4 and $-5$ respectively, then $2p+3q$ is equal to

When you first look at the expression $(1+x{)}^p(1-x{)}^q$, it is easy to feel overwhelmed. You might be tempted to start expanding everything, but the best mathematicians are the laziest ones; they only do the work that is absolutely necessary.

In this problem, we are only interested in the coefficients of $x$ and $x^2$. This means we can completely ignore any terms involving $x^3$ or higher powers. This is the first step to mastering JEE-level problems—identifying what truly matters.

Let us start by expanding our two binomials using the binomial theorem:

Now, we multiply these two series:

To find the coefficient of $x$, we look for combinations that result in $x$. We get $1\cdot (-qx)$ and $px\cdot 1$. Adding these gives $(p-q)x$.

Since the problem states this coefficient is $4$, we have our first elegant equation:

For the $x^2$ term, we collect the products that result in $x^2$:

Summing these gives us the coefficient:

Now, let us simplify the expression for the $x^2$ coefficient. Multiplying by $2$ to clear the denominator, we get:

Rearranging this, we see the structure:

This is where the magic happens! We recognize the perfect square $(p-q{)}^2-(p+q)=-10$. Since we already know $p-q=4$, we substitute this value:

We now have a simple system of linear equations:

Adding them gives $2p=30$, so $p=15$. Subtracting them gives $2q=22$, so $q=11$.

Finally, the question asks for $2p+3q$. Substituting our values:

The final answer is 63. By being selective and using algebraic identities, we turned a potentially messy problem into a clean, logical victory.