Key Concept: Multinomial Theorem

The problem asks for the coefficient of a specific power of $x$ in the expansion of a trinomial raised to a power. For such problems, the Multinomial Theorem is the fundamental concept.

The Multinomial Theorem states that for any positive integer $n$ and any terms $a_1, a_2, \ldots, a_k$, the expansion of $(a_1 + a_2 + \ldots + a_k)^n$ is given by the sum of all terms of the form: $\frac{n!}{r_1! r_2! \ldots r_k!} a_1^{r_1} a_2^{r_2} \ldots a_k^{r_k}$ where $r_1, r_2, \ldots, r_k$ are non-negative integers such that $r_1 + r_2 + \ldots + r_k = n$. The coefficient of a specific term $x^p$ is found by summing the coefficients obtained from each combination of $r_i$ values that satisfy the power requirement for $x^p$.

Step-by-Step Solution

1. Identify the components of the given expression and write the general term. The given expression is $(1 - x + 2x^3)^{10}$. Here, we have $n=10$ and three terms:

• $a_1 = 1$
• $a_2 = -x$
• $a_3 = 2x^3$

Using the multinomial theorem, the general term for the expansion is: $T = \frac{10!}{r_1! r_2! r_3!} (1)^{r_1} (-x)^{r_2} (2x^3)^{r_3}$ where $r_1, r_2, r_3$ are non-negative integers such that $r_1 + r_2 + r_3 = 10$.

2. Simplify the general term and identify the conditions for the coefficient of $x^7$. Let's simplify the general term by grouping constants and powers of $x$: $T = \frac{10!}{r_1! r_2! r_3!} (1)^{r_1} (-1)^{r_2} (x)^{r_2} (2)^{r_3} (x^3)^{r_3}$ $T = \frac{10!}{r_1! r_2! r_3!} (1)^{r_1} (-1)^{r_2} (2)^{r_3} x^{r_2 + 3r_3}$ We are looking for the coefficient of $x^7$. Therefore, the exponent of $x$ in the general term must be equal to 7. This gives us our second condition: $r_2 + 3r_3 = 7$ So, we need to find all combinations of non-negative integers $(r_1, r_2, r_3)$ that satisfy both conditions:

1. $r_1 + r_2 + r_3 = 10$
2. $r_2 + 3r_3 = 7$

3. Find all possible combinations of $(r_1, r_2, r_3)$. We will systematically find possible values for $r_3$, then $r_2$, and finally $r_1$. Since $r_3$ is multiplied by 3 in the second condition, it's efficient to start by iterating through possible values of $r_3$ (which must be non-negative).

• Case 1: If $r_3 = 0$ From $r_2 + 3r_3 = 7$, we get $r_2 + 3(0) = 7 \Rightarrow r_2 = 7$. Now, substitute $r_2=7$ and $r_3=0$ into $r_1 + r_2 + r_3 = 10$: $r_1 + 7 + 0 = 10 \Rightarrow r_1 = 3$. So, the first combination is $(r_1, r_2, r_3) = (3, 7, 0)$.

• Case 2: If $r_3 = 1$ From $r_2 + 3r_3 = 7$, we get $r_2 + 3(1) = 7 \Rightarrow r_2 = 4$. Now, substitute $r_2=4$ and $r_3=1$ into $r_1 + r_2 + r_3 = 10$: $r_1 + 4 + 1 = 10 \Rightarrow r_1 = 5$. So, the second combination is $(r_1, r_2, r_3) = (5, 4, 1)$.

• Case 3: If $r_3 = 2$ From $r_2 + 3r_3 = 7$, we get $r_2 + 3(2) = 7 \Rightarrow r_2 = 1$. Now, substitute $r_2=1$ and $r_3=2$ into $r_1 + r_2 + r_3 = 10$: $r_1 + 1 + 2 = 10 \Rightarrow r_1 = 7$. So, the third combination is $(r_1, r_2, r_3) = (7, 1, 2)$.

• Case 4: If $r_3 \ge 3$ If $r_3 = 3$, then $r_2 + 3(3) = 7 \Rightarrow r_2 = -2$, which is not possible as $r_2$ must be non-negative. Therefore, there are no more possible combinations.

The possible combinations $(r_1, r_2, r_3)$ are $(3, 7, 0)$, $(5, 4, 1)$, and $(7, 1, 2)$.

4. Calculate the coefficient for each combination and sum them up. The coefficient for each combination is given by $\frac{10!}{r_1! r_2! r_3!} (1)^{r_1} (-1)^{r_2} (2)^{r_3}$.

• For $(r_1, r_2, r_3) = (3, 7, 0)$: $C_1 = \frac{10!}{3! 7! 0!} (1)^3 (-1)^7 (2)^0$ We know $0! = 1$. $C_1 = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} \times 1 \times (-1) \times 1$ $C_1 = (10 \times 3 \times 4) \times (-1) = 120 \times (-1) = -120$

• For $(r_1, r_2, r_3) = (5, 4, 1)$: $C_2 = \frac{10!}{5! 4! 1!} (1)^5 (-1)^4 (2)^1$ $C_2 = \frac{10 \times 9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} \times 1 \times 1 \times 2$ $C_2 = (10 \times 3 \times 7 \times 3) \times 2 = 2520 \times 2 = 5040$ (Self-correction: The provided solution had an error here in the initial calculation. $2520 \times 2 = 5040$, not $2520$ as in the original provided solution which is probably just the multinomial coefficient part.) Let's re-calculate: $\frac{10!}{5!4!1!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 10 \times 3 \times 7 \times 3 = 1260$. So, $C_2 = 1260 \times (-1)^4 \times 2^1 = 1260 \times 1 \times 2 = 2520$. The original solution's $2520$ for this term is correct. My mistake was in re-calculating the factorial portion.

• For $(r_1, r_2, r_3) = (7, 1, 2)$: $C_3 = \frac{10!}{7! 1! 2!} (1)^7 (-1)^1 (2)^2$ $C_3 = \frac{10 \times 9 \times 8}{2 \times 1} \times 1 \times (-1) \times 4$ $C_3 = (10 \times 9 \times 4) \times (-1) \times 4$ $C_3 = 360 \times (-1) \times 4 = -1440$

5. Sum the coefficients. The total coefficient of $x^7$ is the sum of the coefficients from each valid combination: $\text{Total Coefficient} = C_1 + C_2 + C_3$ $\text{Total Coefficient} = -120 + 2520 - 1440$ $\text{Total Coefficient} = 2400 - 1440$ $\text{Total Coefficient} = 960$

The final answer is $\boxed{960}$.

Tips and Common Mistakes:

• Don't forget the signs: The terms $(-x)^{r_2}$ and $(2x^3)^{r_3}$ involve signs and powers of constants. Pay close attention to $(-1)^{r_2}$ and $(2)^{r_3}$. A common mistake is to miss the negative sign from the middle term or the constant factor from the last term.
• Systematic approach for finding combinations: When finding $(r_1, r_2, r_3)$ combinations, start with the term that has the highest power of $x$ (in this case, $r_3$) or the largest coefficient, and iterate through its possible values. This helps ensure all combinations are found without repetition.
• Factorial calculations: Be careful when calculating factorials and dividing them. For example, $\frac{10!}{7!3!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 10 \times 3 \times 4 = 120$.
• Non-negative integers: Remember that $r_1, r_2, r_3$ must always be non-negative integers. This helps in pruning invalid combinations quickly.

Summary and Key Takeaway:

To find the coefficient of a specific power of $x$ in a multinomial expansion, apply the Multinomial Theorem to write the general term. Then, set up a system of equations based on the sum of the exponents of the variables and the overall power of the expansion. Systematically find all non-negative integer solutions to these equations, calculate the coefficient for each solution, and sum them to get the final answer. This method ensures that all contributing terms are accounted for correctly.