Detailed Solution for the Binomial Expansion Problem

1. Key Concept: The Multinomial Theorem

This problem revolves around the expansion of a trinomial raised to a power, which is best handled using the Multinomial Theorem. For an expression of the form $(x_1 + x_2 + \dots + x_m)^n$, the general term in its expansion is given by: $T = \frac{n!}{k_1!k_2!\dots k_m!} x_1^{k_1} x_2^{k_2} \dots x_m^{k_m}$ where $k_1, k_2, \dots, k_m$ are non-negative integers such that $k_1 + k_2 + \dots + k_m = n$.

In our problem, we have $(a+bx+cx^2)^{10}$. Here, $n=10$, and the terms are $x_1 = a$, $x_2 = bx$, and $x_3 = cx^2$. So, the general term for this expansion becomes: $T = \frac{10!}{k_1!k_2!k_3!} (a)^{k_1} (bx)^{k_2} (cx^2)^{k_3}$ To simplify and identify the power of $x$, we group the coefficients and the powers of $x$: $T = \frac{10!}{k_1!k_2!k_3!} a^{k_1} b^{k_2} c^{k_3} x^{k_2 + 2k_3}$ where $k_1 + k_2 + k_3 = 10$.

2. Finding the Coefficient of $x^1$ ($p_1$)

To find the coefficient of $x^1$, we need the exponent of $x$ in the general term to be $1$. So, we set the condition: $k_2 + 2k_3 = 1$ Since $k_2$ and $k_3$ must be non-negative integers:

• If $k_3 = 0$, then $k_2 = 1$.
  • Knowing $k_1 + k_2 + k_3 = 10$, we can find $k_1$: $k_1 = 10 - 1 - 0 = 9$.
  • This gives us the triplet $(k_1, k_2, k_3) = (9, 1, 0)$. This is the only combination that results in $x^1$.

Now, we substitute these values into the general term's coefficient part: $p_1 = \frac{10!}{9!1!0!} a^9 b^1 c^0$ Recall that $0! = 1$. $p_1 = \frac{10 \times 9!}{9! \times 1 \times 1} a^9 b = 10a^9 b$ We are given that $p_1 = 20$. Therefore: $10a^9 b = 20$ $a^9 b = 2 \quad (\text{Equation 1})$

3. Finding the Coefficient of $x^2$ ($p_2$)

Next, we need to find the coefficient of $x^2$. This means the exponent of $x$ must be $2$: $k_2 + 2k_3 = 2$ Again, considering non-negative integer values for $k_2$ and $k_3$, we have two possible cases:

• Case 1: If $k_3 = 0$, then $k_2 = 2$.

  • For this case, $k_1 = 10 - k_2 - k_3 = 10 - 2 - 0 = 8$.
  • The triplet is $(k_1, k_2, k_3) = (8, 2, 0)$.
  • The coefficient from this case is: $\frac{10!}{8!2!0!} a^8 b^2 c^0 = \frac{10 \times 9}{2 \times 1} a^8 b^2 = 45a^8 b^2$

• Case 2: If $k_3 = 1$, then $k_2 = 0$.

  • For this case, $k_1 = 10 - k_2 - k_3 = 10 - 0 - 1 = 9$.
  • The triplet is $(k_1, k_2, k_3) = (9, 0, 1)$.
  • The coefficient from this case is: $\frac{10!}{9!0!1!} a^9 b^0 c^1 = \frac{10 \times 9!}{9! \times 1 \times 1} a^9 c = 10a^9 c$

The total coefficient $p_2$ is the sum of coefficients from all possible cases: $p_2 = 45a^8 b^2 + 10a^9 c$ We are given that $p_2 = 210$. Therefore: $45a^8 b^2 + 10a^9 c = 210 \quad (\text{Equation 2})$

4. Solving for $a, b, c$

We now have a system of two equations with three variables $a, b, c$:

1. $a^9 b = 2$
2. $45a^8 b^2 + 10a^9 c = 210$

We are given that $a, b, c \in \mathbb{N}$ (natural numbers, i.e., positive integers). This constraint is crucial for solving the problem efficiently.

Let's analyze Equation 1: $a^9 b = 2$. Since $a$ and $b$ must be positive integers, the only possible integer values for $a$ are $1$.

• If $a=1$, then $(1)^9 b = 2 \implies b = 2$.
• If $a$ were any other integer, $a^9$ would be too large, or $b$ would not be an integer. For instance, if $a=2$, then $2^9 b = 2 \implies b = 1/2$, which is not a natural number.

Thus, the only valid integer solution for $(a, b)$ is $a=1$ and $b=2$.

Now, substitute $a=1$ and $b=2$ into Equation 2 to find $c$: $45(1)^8 (2)^2 + 10(1)^9 c = 210$ $45(1)(4) + 10(1)c = 210$ $180 + 10c = 210$ Subtract $180$ from both sides: $10c = 210 - 180$ $10c = 30$ Divide by $10$: $c = 3$ Since $c=3$ is a natural number, our values $a=1, b=2, c=3$ are consistent with all given conditions.

5. Final Calculation: Finding $2(a+b+c)$

The problem asks for the value of $2(a+b+c)$. Substitute the found values of $a, b, c$: $2(a+b+c) = 2(1 + 2 + 3)$ $= 2(6)$ $= 12$

Tips and Common Mistakes to Avoid:

• Understanding the Multinomial Theorem: Ensure you correctly identify $n$, $x_i$, and the conditions for $k_i$. A common mistake is misinterpreting the exponents of $x$ (e.g., forgetting the $x^2$ in $cx^2$).
• Checking all combinations: When finding coefficients like $p_2$, make sure you enumerate all possible non-negative integer combinations of $k_2$ and $k_3$ that satisfy the exponent condition ($k_2 + 2k_3 = 2$).
• Using Constraints: The constraint $a, b, c \in \mathbb{N}$ (natural numbers/positive integers) is extremely important. Without it, the problem would be significantly harder or have multiple solutions. Always look for such constraints.
• Factorial Calculations: Be careful with factorial simplifications, e.g., $\frac{10!}{9!1!0!} = \frac{10 \times 9!}{9! \times 1 \times 1} = 10$.

Summary:

This problem is a good test of applying the Multinomial Theorem to find specific coefficients in a polynomial expansion. The key steps involved setting up the general term, identifying the conditions for the powers of $x$, forming a system of equations from the given coefficients, and solving these equations by carefully utilizing the constraint that $a, b, c$ are natural numbers. The final calculation is straightforward once $a, b, c$ are determined.