Question

The coefficient of $x^{\frac{n^2+n-14}{2}}$ in $(x-1)(x^2-2)(x^3-3)(x^4-4)...(x^n-n)$, $n \geq 30$ is equal to

Answer 1

13

Answer 2

The given expression is $P(x) = (x-1)(x^2-2)(x^3-3)(x^4-4)...(x^n-n)$. We need to find the coefficient of $x^{\frac{n^2+n-14}{2}}$.

Let's consider a general term in the expansion of $P(x)$. Each term is formed by picking either $x^k$ or $-k$ from each factor $(x^k-k)$. Let $S$ be the set of indices $k$ for which we choose $x^k$. Let $S'$ be the set of indices $k$ for which we choose $-k$. Then $S \cup S' = \{1, 2, ..., n\}$ and $S \cap S' = \emptyset$.

The power of $x$ in such a term will be $\sum_{k \in S} k$. The coefficient of this term will be $\prod_{k \in S'} (-k)$.

The maximum possible power of $x$ in the product is when we pick $x^k$ from every factor, which is $\sum_{k=1}^n k = \frac{n(n+1)}{2}$.

The given power of $x$ is $\frac{n^2+n-14}{2}$. We can write this as $\frac{n(n+1)}{2} - 7$.

The power of $x$ in a general term is $\sum_{k \in S} k$. Also, $\sum_{k \in S} k = \left(\sum_{k=1}^n k\right) - \left(\sum_{k \in S'} k\right)$. Comparing this with the given power: $\frac{n(n+1)}{2} - \sum_{k \in S'} k = \frac{n(n+1)}{2} - 7$. This implies that $\sum_{k \in S'} k = 7$.

We need to find all possible sets $S'$ of distinct positive integers whose sum is 7. Since $n \geq 30$, all integers from 1 to 7 are available as indices.

Let's list the possible sets $S'$ and the corresponding coefficients:

1. $S'$ contains one element: The only possibility is $S' = \{7\}$. The coefficient is $(-1)^1 \cdot 7 = -7$.

2. $S'$ contains two distinct elements: Let the elements be $k_1, k_2$ such that $k_1 + k_2 = 7$ and $k_1 \neq k_2$. Possible sets:

   • $S' = \{1, 6\}$. The coefficient is $(-1)^2 \cdot (1 \cdot 6) = 6$.
   • $S' = \{2, 5\}$. The coefficient is $(-1)^2 \cdot (2 \cdot 5) = 10$.
   • $S' = \{3, 4\}$. The coefficient is $(-1)^2 \cdot (3 \cdot 4) = 12$.

3. $S'$ contains three distinct elements: Let the elements be $k_1, k_2, k_3$ such that $k_1 + k_2 + k_3 = 7$ and $k_1, k_2, k_3$ are distinct. The only possibility is $S' = \{1, 2, 4\}$ (since $1+2+3=6$, the next smallest is $1+2+4=7$). The coefficient is $(-1)^3 \cdot (1 \cdot 2 \cdot 4) = -8$.

4. $S'$ contains four or more distinct elements: The smallest sum of four distinct positive integers is $1+2+3+4 = 10$, which is greater than 7. Therefore, no sets with four or more elements are possible.

To find the total coefficient of $x^{\frac{n^2+n-14}{2}}$, we sum the coefficients from all valid cases: Total coefficient = $(-7) + 6 + 10 + 12 + (-8)$ Total coefficient = $-7 + 6 + 10 + 12 - 8$ Total coefficient = $-1 + 10 + 12 - 8$ Total coefficient = $9 + 12 - 8$ Total coefficient = $21 - 8$ Total coefficient = $13$.

The final answer is $\boxed{13}$.

Explanation of the solution:

The problem asks for the coefficient of a specific power of $x$ in the expansion of a product of terms $(x^k - k)$. Each term in the expanded polynomial is formed by choosing either $x^k$ or $-k$ from each factor $(x^k-k)$. The maximum possible power of $x$ is $\sum_{k=1}^n k = \frac{n(n+1)}{2}$. The target power of $x$ is $\frac{n(n+1)}{2} - 7$. This means that from the factors $(x^k - k)$, we must have chosen $-k$ for a set of indices $S'$ such that their sum $\sum_{k \in S'} k = 7$. The coefficient for such a term is $\prod_{k \in S'} (-k)$. We enumerate all distinct sets $S'$ of positive integers that sum to 7:

1. $\{7\}$: Coefficient $= (-7)$
2. $\{1, 6\}$: Coefficient $= (-1)(-6) = 6$
3. $\{2, 5\}$: Coefficient $= (-2)(-5) = 10$
4. $\{3, 4\}$: Coefficient $= (-3)(-4) = 12$
5. $\{1, 2, 4\}$: Coefficient $= (-1)(-2)(-4) = -8$ The sum of these coefficients is $-7 + 6 + 10 + 12 - 8 = 13$.

Answer:

The coefficient of $x^{\frac{n^2+n-14}{2}}$ is $13$. The option is not given, but the calculated value is 13.