Animated Solution for Mathematics - Binomial Theorem: Let $(a+bx+c{x}^2{)}^{10}={\sum }_{i=0}^{20}{p}_i{x}^i,a,b,c\in \mathbb{N}$. If $p_1=20$ and $p_2=210$, then $2(a+b+c)$ is equal to

Welcome, future engineer. Today, we are not just solving a problem; we are peeling back the layers of a mathematical structure. When you see an expression like $(a+bx+c{x}^2{)}^{10}$, your first instinct might be to panic.

It looks like a mess of variables. But I want you to see it differently. Imagine this expression as a machine; you feed it indices, and it spits out terms. Our job is to reverse-engineer the machine to find the values of $a$, $b$, and $c$.

The Heart of the Beast

In any expansion of the form $(x_1+x_2+x_3{)}^n$, the general term is governed by the Multinomial Theorem. It is defined as:

In our specific case, where we have $(a+bx+c{x}^2{)}^{10}$, the general term becomes:

Here, the constraints are simple but absolute: $r_1+r_2+r_3=10$, and $r_1,r_2,r_3$ must be non-negative integers. If you look closely at the term, you will see that the power of $x$ is determined by $r_2+2r_3$. This is the key to the entire problem and acts as our filter.

We are given that $p_1=20$. This means the coefficient of $x^1$ is $20$. To find this, we set our filter to $1$:

Since $r_2$ and $r_3$ are non-negative integers, the only way to satisfy this is $r_3=0$ and $r_2=1$. This automatically forces $r_1=10-1-0=9$.

Now, we substitute these into our general term formula:

We are told this equals $20$. So, $10a^{9}b=20$, which simplifies to $a^{9}b=2$. Because $a$ and $b$ are natural numbers, we have no choice but to accept $a=1$ and $b=2$.

Now, the plot thickens. We are given $p_2=210$, which is the coefficient of $x^2$. We set our filter to $2$:

This time, the universe offers us two paths. We must explore both:

1. Case 1: $r_3=0$. Then $r_2=2$. This implies $r_1=10-2-0=8$. 2. Case 2: $r_3=1$. Then $r_2=0$. This implies $r_1=10-0-1=9$.

We must calculate the contribution from both cases and add them together. For Case 1, the term is $\frac{10!}{8!2!0!}{a}^8{b}^2=45a^8{b}^2$. For Case 2, the term is $\frac{10!}{9!0!1!}{a}^{9}c=10a^{9}c$.

So, the total coefficient $p_2$ is:

We are almost at the finish line. We know $a=1$, $b=2$, and $p_2=210$. Let us substitute these values into our equation:

Subtracting $180$ from both sides gives us $30=10c$, which leads us to the beautiful conclusion that $c=3$. We have found our trio: $a=1,b=2,c=3$.

The question asks for $2(a+b+c)$:

And there it is. The complexity of the multinomial expansion dissolves when you approach it with systematic logic. Never fear the variables; they are just placeholders waiting for you to define them.