Animated Solution for Mathematics - Binomial Theorem: If the coefficient of $x$ in the expansion of $(a{x}^2+bx+c)(1-2x{)}^{26}$ is -56 and the coefficients of $x^2$ and $x^3$ are both zero, then $a+b+c$ is equal to :

Welcome, future engineers! Today, we are going to dismantle a problem that, at first glance, looks like a nightmare of arithmetic. We are faced with the product of a quadratic, $(a{x}^2+bx+c)$, and a binomial raised to the power of twenty-six, $(1-2x{)}^{26}$.

The sheer magnitude of that exponent, twenty-six, is designed to make you panic. It is a classic JEE trap. But here is the secret: you do not need to expand the entire binomial.

You only need to be a sniper, not a machine gunner. We only care about the coefficients of $x$, $x^2$, and $x^3$. Anything beyond that is irrelevant noise. Let us begin our journey.

First, let us invoke the Binomial Theorem. We know that:

In our case, $y=-2x$ and $n=26$. We only need the first four terms because our quadratic only goes up to $x^2$.

If we multiply $a{x}^2$ by anything higher than $x^1$ from the binomial, we get $x^3$ or higher. So, let us calculate the first few terms:

These four values are our ammunition.

Now, we play the matching game. We are multiplying $(a{x}^2+bx+c)$ by $(1-52x+1300x^2-20800x^3+\dots )$.

To find the coefficient of $x$, we look for combinations that produce $x$. That is $c\times (-52x)$ and $bx\times 1$. This gives us:

That is our first equation. For $x^2$, we need $c\times (1300x^2)$, $bx\times (-52x)$, and $a{x}^2\times 1$. This yields:

Finally, for $x^3$, we combine $c\times (-20800x^3)$, $bx\times (1300x^2)$, and $a{x}^2\times (-52x)$, resulting in:

Look at that third equation. It looks terrifying, but notice the pattern? Every single term is a multiple of 52.

If we divide the entire equation by $-52$, it simplifies beautifully to:

Now, we have a clean system of linear equations. By subtracting this simplified equation from our second equation, the variable $a$ vanishes, leaving us with a direct relationship between $b$ and $c$.

We find that $27b=900c$, or:

Substituting this into our first equation, we solve for $c$ and find $c=3$. From there, the dominoes fall: $b=100$ and $a=1300$.

The sum $a+b+c=1403$. We have conquered the monster by simply refusing to be intimidated by its size. Keep this mindset, and you will solve any problem the JEE throws at you.