Key Concept: The Binomial Theorem

The core concept for solving this problem is the Binomial Theorem, which provides a formula for expanding binomials raised to any non-negative integer power. For an expression of the form $(a+b)^n$, the expansion is given by: $(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$ where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ is the binomial coefficient. In this problem, we are dealing with expressions of the form $(1+y)^n$. For these, the expansion simplifies to: $(1+y)^n = \binom{n}{0} + \binom{n}{1}y + \binom{n}{2}y^2 + \dots + \binom{n}{n}y^n$ Our goal is to find the coefficient of $x^{10}$ in the product of three such expansions. This means we need to identify terms from each individual expansion whose powers of $x$ sum up to $10$.

Step-by-Step Expansion of Each Factor

First, we expand each of the three factors using the Binomial Theorem to identify the possible powers of $x$ and their corresponding coefficients.

1. Expansion of $(1+x)^2$: Here, $n=2$ and $y=x$. $(1+x)^2 = \binom{2}{0}x^0 + \binom{2}{1}x^1 + \binom{2}{2}x^2$ $(1+x)^2 = 1 \cdot 1 + 2 \cdot x + 1 \cdot x^2 = 1 + 2x + x^2$ The possible terms are $x^0$ (coefficient 1), $x^1$ (coefficient 2), and $x^2$ (coefficient 1).

2. Expansion of $(1+x^2)^3$: Here, $n=3$ and $y=x^2$. $(1+x^2)^3 = \binom{3}{0}(x^2)^0 + \binom{3}{1}(x^2)^1 + \binom{3}{2}(x^2)^2 + \binom{3}{3}(x^2)^3$ $(1+x^2)^3 = 1 \cdot 1 + 3 \cdot x^2 + 3 \cdot x^4 + 1 \cdot x^6 = 1 + 3x^2 + 3x^4 + x^6$ The possible terms are $x^0$ (coefficient 1), $x^2$ (coefficient 3), $x^4$ (coefficient 3), and $x^6$ (coefficient 1).

3. Expansion of $(1+x^3)^4$: Here, $n=4$ and $y=x^3$. $(1+x^3)^4 = \binom{4}{0}(x^3)^0 + \binom{4}{1}(x^3)^1 + \binom{4}{2}(x^3)^2 + \binom{4}{3}(x^3)^3 + \binom{4}{4}(x^3)^4$ $(1+x^3)^4 = 1 \cdot 1 + 4 \cdot x^3 + 6 \cdot x^6 + 4 \cdot x^9 + 1 \cdot x^{12} = 1 + 4x^3 + 6x^6 + 4x^9 + x^{12}$ The possible terms are $x^0$ (coefficient 1), $x^3$ (coefficient 4), $x^6$ (coefficient 6), $x^9$ (coefficient 4), and $x^{12}$ (coefficient 1).

Identifying Combinations for $x^{10}$

We need to find combinations of terms, one from each expanded factor, such that the sum of their powers of $x$ equals $10$. Let $x^{a_1}$ be a term from $(1+x)^2$, $x^{a_2}$ from $(1+x^2)^3$, and $x^{a_3}$ from $(1+x^3)^4$. We are looking for all sets of $(a_1, a_2, a_3)$ such that $a_1 + a_2 + a_3 = 10$.

We will systematically list the possible powers from each expansion and find combinations that sum to $10$.

• Case 1: Term from $(1+x)^2$ is $x^0$ (coefficient 1). We need $a_2 + a_3 = 10$.

  • If $a_2 = x^4$ (coefficient 3), then $a_3$ must be $x^6$ (coefficient 6). Combination: $x^0 \cdot x^4 \cdot x^6 = x^{10}$. Product of coefficients: $1 \times 3 \times 6 = 18$.
  • (Other possible $a_2$ values like $x^0, x^2, x^6$ would require $a_3$ to be $x^{10}, x^8, x^4$ respectively, which are not available from $(1+x^3)^4$ except for $x^{12}$ which is too high or $x^9, x^6, x^3, x^0$.)

• Case 2: Term from $(1+x)^2$ is $x^1$ (coefficient 2). We need $a_2 + a_3 = 9$.

  • If $a_2 = x^0$ (coefficient 1), then $a_3$ must be $x^9$ (coefficient 4). Combination: $x^1 \cdot x^0 \cdot x^9 = x^{10}$. Product of coefficients: $2 \times 1 \times 4 = 8$.
  • If $a_2 = x^6$ (coefficient 1), then $a_3$ must be $x^3$ (coefficient 4). Combination: $x^1 \cdot x^6 \cdot x^3 = x^{10}$. Product of coefficients: $2 \times 1 \times 4 = 8$.

• Case 3: Term from $(1+x)^2$ is $x^2$ (coefficient 1). We need $a_2 + a_3 = 8$.

  • If $a_2 = x^2$ (coefficient 3), then $a_3$ must be $x^6$ (coefficient 6). Combination: $x^2 \cdot x^2 \cdot x^6 = x^{10}$. Product of coefficients: $1 \times 3 \times 6 = 18$.

Calculating the Total Coefficient

To find the total coefficient of $x^{10}$, we sum the products of coefficients from all the valid combinations identified: Total Coefficient $= 18 + 8 + 8 + 18 = 52$.

Tips for Success & Common Pitfalls

• Systematic Approach: Always list out the terms and their coefficients for each factor. Then, systematically iterate through possible combinations to avoid missing any or duplicating efforts. Starting with the lowest possible power from the first factor and working upwards helps.
• Exponents vs. Powers: Be careful when expanding terms like $(x^2)^k$ or $(x^3)^k$. The resulting power of $x$ is $2k$ or $3k$, not $k$.
• Coefficient vs. Power: Remember to multiply the coefficients of the chosen terms, not their powers, to get the combined coefficient for $x^{10}$.
• Check Bounds: Ensure that the powers you choose from each factor actually exist within its expansion. For example, $(1+x^2)^3$ does not have an $x^1$ term.

Summary

By systematically expanding each binomial factor using the Binomial Theorem and then identifying all combinations of terms (one from each expansion) whose powers of $x$ sum to $10$, we were able to calculate the corresponding product of coefficients for each combination. Summing these products yields the final coefficient of $x^{10}$, which is $52$. This method ensures accuracy and clarity in complex polynomial multiplications.