Key Concept: Binomial Theorem and Coefficient of a Term in a Product

To find the coefficient of a specific power of $x$ in the expansion of a product of binomials, we leverage the Binomial Theorem. The theorem states that for any non-negative integer $n$, the expansion of $(a+b)^n$ is given by: $(a+b)^n = \sum_{k=0}^{n} {}^n C_k a^{n-k} b^k$ In this problem, we will first simplify the base expression into a product of simpler binomials. Then, we will find the general term for each binomial expansion. Finally, we will identify all combinations of terms from these expansions whose product yields the desired power of $x$, and sum their corresponding coefficients.

Step 1: Simplify the Base Expression

The given expression is $(1 + x + x^2 + x^3)^6$. We observe that the terms inside the parenthesis form a finite geometric series or can be factored directly. $1 + x + x^2 + x^3 = 1(1+x) + x^2(1+x)$ $= (1+x)(1+x^2)$ This factorization is crucial as it transforms the problem into finding the coefficient in a product of two simpler binomial expressions. So, the original expression becomes: $(1 + x + x^2 + x^3)^6 = ((1+x)(1+x^2))^6$ $= (1+x)^6 (1+x^2)^6$

Step 2: Apply the Binomial Theorem to Each Factor

Now we have a product of two binomial expansions: $(1+x)^6$ and $(1+x^2)^6$. We need to find the general term for each.

For the first factor, $(1+x)^6$: Using the Binomial Theorem with $a=1$, $b=x$, and $n=6$, the general term $T_{r+1}$ is: $T_{r+1} = {}^6 C_r (1)^{6-r} (x)^r = {}^6 C_r x^r$ where $r$ can range from $0$ to $6$.

For the second factor, $(1+x^2)^6$: Using the Binomial Theorem with $a=1$, $b=x^2$, and $n=6$, the general term $T_{s+1}$ is: $T_{s+1} = {}^6 C_s (1)^{6-s} (x^2)^s = {}^6 C_s x^{2s}$ where $s$ can range from $0$ to $6$. Note: It's important to use different index variables (like $r$ and $s$) for each independent expansion to avoid confusion.

Step 3: Identify Conditions for the Coefficient of $x^4$

We are looking for the coefficient of $x^4$ in the product of these two general terms. The product of the general terms is $({}^6 C_r x^r) \cdot ( {}^6 C_s x^{2s}) = {}^6 C_r {}^6 C_s x^{r+2s}$. For the term to be $x^4$, the powers of $x$ must sum to $4$: $r + 2s = 4$

Step 4: Find All Valid Combinations of $r$ and $s$

We need to find non-negative integer pairs $(r, s)$ that satisfy $r + 2s = 4$, subject to the constraints $0 \le r \le 6$ and $0 \le s \le 6$. We can systematically test values for $s$:

• Case 1: If $s=0$ $r + 2(0) = 4 \implies r = 4$. This pair $(r, s) = (4, 0)$ is valid since $0 \le 4 \le 6$ and $0 \le 0 \le 6$. The coefficient from this case is ${}^6 C_4 \cdot {}^6 C_0$.

• Case 2: If $s=1$ $r + 2(1) = 4 \implies r = 2$. This pair $(r, s) = (2, 1)$ is valid since $0 \le 2 \le 6$ and $0 \le 1 \le 6$. The coefficient from this case is ${}^6 C_2 \cdot {}^6 C_1$.

• Case 3: If $s=2$ $r + 2(2) = 4 \implies r = 0$. This pair $(r, s) = (0, 2)$ is valid since $0 \le 0 \le 6$ and $0 \le 2 \le 6$. The coefficient from this case is ${}^6 C_0 \cdot {}^6 C_2$.

• Case 4: If $s=3$ $r + 2(3) = 4 \implies r = -2$. This is not a valid combination because $r$ must be non-negative. For any $s > 2$, $2s$ will be greater than 4, making $r$ negative. Therefore, we have found all valid combinations.

Step 5: Calculate the Coefficients for Each Combination

Now, we calculate the values for each binomial coefficient ${}^n C_k = \frac{n!}{k!(n-k)!}$:

• ${}^6 C_0 = 1$
• ${}^6 C_1 = 6$
• ${}^6 C_2 = \frac{6 \times 5}{2 \times 1} = 15$
• ${}^6 C_4 = \frac{6 \times 5 \times 4 \times 3}{4 \times 3 \times 2 \times 1} = 15$ (Recall ${}^n C_k = {}^n C_{n-k}$, so ${}^6 C_4 = {}^6 C_{6-4} = {}^6 C_2$)

Now, we calculate the product of coefficients for each valid $(r, s)$ pair:

• From $(r=4, s=0)$: ${}^6 C_4 \cdot {}^6 C_0 = 15 \cdot 1 = 15$
• From $(r=2, s=1)$: ${}^6 C_2 \cdot {}^6 C_1 = 15 \cdot 6 = 90$
• From $(r=0, s=2)$: ${}^6 C_0 \cdot {}^6 C_2 = 1 \cdot 15 = 15$

Step 6: Sum the Coefficients

The total coefficient of $x^4$ is the sum of the coefficients from all valid combinations: $\text{Coefficient of } x^4 = 15 + 90 + 15 = 120$

Tips and Common Mistakes:

1. Simplification First: Always look for opportunities to simplify the base expression, especially if it's a geometric series or can be factored. This often reduces the complexity significantly.
2. Distinct Indices: When multiplying expansions, use different index variables (e.g., $r$ and $s$) for each binomial expansion. Using the same variable for both can lead to errors.
3. Boundary Conditions: Remember that the index $r$ (or $s$) in ${}^n C_r$ must be an integer between $0$ and $n$ (inclusive). Always check these bounds when finding combinations.
4. Combinations, Not Just Terms: When looking for a specific power, remember to sum the coefficients from ALL possible ways that power can be formed.

Summary and Key Takeaway:

The problem effectively tests the understanding of the Binomial Theorem and the ability to handle products of polynomial expansions. The key steps involve simplifying the given expression, applying the Binomial Theorem to obtain general terms for each factor, setting up an equation for the powers, systematically finding all valid integer solutions for the indices, calculating the corresponding binomial coefficients, and finally summing them up. This method ensures that all possible contributions to the target power of $x$ are accounted for.