Finding the Coefficient of x⁹⁹

The problem asks for the coefficient of the \( x^{99} \) term in the expansion of the polynomial product \( (x - 1)(x - 2)(x - 3) \cdots (x - 100) \). This is a polynomial of degree 100.

Let the given polynomial be \( P(x) = (x - 1)(x - 2)(x - 3) \cdots (x - 100) \). We can express this polynomial in the standard form:

\( P(x) = a_{100}x^{100} + a_{99}x^{99} + a_{98}x^{98} + \cdots + a_1x + a_0 \)

We are interested in finding the value of the coefficient \( a_{99} \).

Understanding Polynomial Expansion from Roots

Consider a general polynomial that is factored in terms of its roots \( r_1, r_2, \dots, r_n \):

\( P(x) = (x - r_1)(x - r_2) \cdots (x - r_n) \)

When this polynomial is expanded, the terms are formed by choosing either \( x \) or the constant term from each factor \( (x - r_i) \).

• To get the highest power, \( x^n \), we choose \( x \) from all \( n \) factors. The coefficient is 1.
• To get the next highest power, \( x^{n-1} \), we need to choose \( x \) from \( n-1 \) factors and the constant term \( -r_i \) from one factor. We sum up all possible ways to do this. For example, for \( (x-r_1)(x-r_2)(x-r_3) \), the \( x^2 \) term comes from \( x \cdot x \cdot (-r_3) \), \( x \cdot (-r_2) \cdot x \), and \( (-r_1) \cdot x \cdot x \), which sums to \( (-r_3 - r_2 - r_1)x^2 = -(r_1+r_2+r_3)x^2 \).
• In general, the coefficient of \( x^{n-1} \) in the expansion of \( (x - r_1)(x - r_2) \cdots (x - r_n) \) is \( -(r_1 + r_2 + \cdots + r_n) \).

Applying the Concept to the Given Problem

In our problem, the polynomial is \( (x - 1)(x - 2)(x - 3) \cdots (x - 100) \). This matches the form \( (x - r_1)(x - r_2) \cdots (x - r_n) \) where:

• The degree of the polynomial is \( n = 100 \).
• The roots are \( r_1 = 1, r_2 = 2, r_3 = 3, \dots, r_{100} = 100 \).

We are looking for the coefficient of \( x^{99} \). This corresponds to the term with power \( x^{n-1} \) since \( 99 = 100 - 1 \).

Based on the general form, the coefficient of \( x^{99} \) is \( -(r_1 + r_2 + r_3 + \cdots + r_{100}) \).

Substituting the roots from our problem:

Coefficient of \( x^{99} = -(1 + 2 + 3 + \cdots + 100) \)

Calculating the Sum of the Roots

The sum \( 1 + 2 + 3 + \cdots + 100 \) is the sum of the first 100 positive integers. The formula for the sum of the first \( k \) positive integers is \( S_k = \frac{k(k+1)}{2} \).

Here, \( k = 100 \). So, the sum is:

\( 1 + 2 + 3 + \cdots + 100 = \frac{100(100 + 1)}{2} = \frac{100 \times 101}{2} \)

\( = \frac{10100}{2} = 5050 \)

Final Coefficient of x⁹⁹

The coefficient of \( x^{99} \) is the negative of this sum:

Coefficient of \( x^{99} = -(1 + 2 + 3 + \cdots + 100) = -5050 \)

Step-by-Step Calculation

1. Identify the polynomial and the coefficient needed: We need the coefficient of \( x^{99} \) in \( (x - 1)(x - 2) \cdots (x - 100) \).
2. Recognize the form: The polynomial is in the form \( (x - r_1)(x - r_2) \cdots (x - r_n) \), where \( n=100 \) and roots are \( r_i = i \) for \( i=1, \dots, 100 \).
3. Recall the coefficient formula: The coefficient of \( x^{n-1} \) in such an expansion is the negative sum of the roots, \( -(r_1 + r_2 + \cdots + r_n) \).
4. Apply the formula: The coefficient of \( x^{99} \) is \( -(1 + 2 + \cdots + 100) \).
5. Calculate the sum: Use the formula for the sum of the first 100 integers: \( \frac{100(101)}{2} = 5050 \).
6. Determine the coefficient: The coefficient is \( -5050 \).

The calculated coefficient of \( x^{99} \) in the expansion of \( (x - 1)(x - 2)(x - 3) \cdots (x - 100) \) is \( -5050 \).

Revision Table - Key Concepts

Additional Information - Vieta's Formulas

Vieta's formulas provide relationships between the coefficients of a polynomial and the sums and products of its roots. For a polynomial \( a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0 \), with roots \( r_1, r_2, \dots, r_n \), the following relationships hold (assuming \( a_n = 1 \)):

• Sum of roots: \( r_1 + r_2 + \cdots + r_n = -\frac{a_{n-1}}{a_n} \)
• Sum of products of roots taken two at a time: \( \sum_{i<j} r_i r_j = \frac{a_{n-2}}{a_n} \)
• Sum of products of roots taken three at a time: \( \sum_{i<j<k} r_i r_j r_k = -\frac{a_{n-3}}{a_n} \)
• ...
• Product of roots: \( r_1 r_2 \cdots r_n = (-1)^n \frac{a_0}{a_n} \)

In our problem, \( P(x) = (x - 1)(x - 2) \cdots (x - 100) \). If we expand this, the leading coefficient \( a_{100} \) is 1. The roots are \( r_i = i \) for \( i=1, \dots, 100 \).

The coefficient of \( x^{99} \) is \( a_{99} \). According to Vieta's formulas:

\( r_1 + r_2 + \cdots + r_{100} = -\frac{a_{99}}{a_{100}} \)

Since \( a_{100} = 1 \), this simplifies to:

\( r_1 + r_2 + \cdots + r_{100} = -a_{99} \)

Therefore, \( a_{99} = -(r_1 + r_2 + \cdots + r_{100}) \).

This confirms our approach that the coefficient of \( x^{99} \) is the negative of the sum of the roots, which are 1, 2, ..., 100.

The calculation \( -(1 + 2 + \cdots + 100) = -5050 \) gives the coefficient of \( x^{99} \).