Problem 80 - A Complete Resource Book in Mathematics for JEE Main

Question

The coefficient of \(x^{99}\) and \(x^{98}\) in the polynomial \((x-1)(x-2)(x-3) \ldots(x-100)\) are (A) \(-5050\) and 12482075 (B) \(-4050\) and 12582075 (C) \(-5050\) and 12582075 (D) None of these

Step-by-Step Solution

Answer

The coefficients are (A) \(-5050\) and 12482075.

Step 1: Understand the Polynomial

The given expression \((x-1)(x-2)(x-3) \ldots (x-100)\) is a polynomial of degree 100, so it can be expanded into a sum of terms, with each term having a certain coefficient. We want the coefficients of \(x^{99}\) and \(x^{98}\).

Step 2: Find the Coefficient of \(x^{99}\)

The coefficient of \(x^{99}\) is found by considering contributions from the terms where we choose one constant to omit in the multiplication process. This is equivalent to summing the numbers from 1 to 100. In terms of formula, this is given by the negative sum, calculated as: \(- (1 + 2 + 3 + \ldots + 100) = - \frac{100(101)}{2} = -5050\).

Step 3: Find the Coefficient of \(x^{98}\)

The coefficient of \(x^{98}\) is found by choosing two different constants omitted in the product. This can be expressed using combinations:\(\sum_{1 \leq i < j \leq 100} ij\). When simplified, this results in a known formula: \[- \frac{(1+2+\ldots+100)^2 - (1^2 + 2^2 + \ldots + 100^2)}{2}= 12482075\].

Step 4: Match with the Given Options

Now compare the calculated coefficients of \(-5050\) for \(x^{99}\) and \(12482075\) for \(x^{98}\) with the options provided. Option (A) matches the coefficients.

Key Concepts

Polynomial Expansion

Polynomial expansion involves expressing a polynomial in its extended form. When you have a polynomial like \((x-1)(x-2)(x-3) \ldots (x-100)\), it is initially expressed in its factored form. This polynomial is a degree 100 polynomial because it has 100 linear factors each with variable \(x\). If you multiply these terms out completely, you'll get an expanded polynomial expression with terms ranging from \(x^{100}\) down to the constant term.The magic of polynomial expansion comes into play when you realize that each term in the expanded form can be identified by the number of \(x\) variables being multiplied together. For a \(k\)-th term, it is characterized by a combination of the original variables and numerics from the constants in the brackets. Expanding the polynomial helps in identifying terms like \(x^{99}\) and \(x^{98}\) as seen in your exercise.

Combinatorial Coefficient Calculation

Combinatorial coefficient calculation is essentially the idea of determining the coefficients in front of each term in an expanded polynomial. For the term \(x^{99}\), one constant needs to be omitted from each multiplication group. To get its coefficient, calculate the sum of all integers from 1 to 100, but here's the kicker: the coefficient is the sum taken negatively, because omitting one number leaves all others to multiply, capturing a situation where missing one affects the sum negatively.For \(x^{98}\), things get trickier as you remove two numbers at a time from the multiplication group, effectively calculating the combinations of choosing 2 numbers out of 100 and multiplying them as combinations \(\binom{n}{2}\). The combinatorial coefficient calculation is handy especially in identifying specific terms without fully expanding the polynomial!

Integer Sum Formula

The integer sum formula is a straightforward yet powerful mathematical tool. It's used to find the sum of a sequence of integers. The specific formula \(1 + 2 + 3 + \ldots + n = \frac{n(n+1)}{2}\) applies when summing up integers from 1 to \(n\). In your exercise, this is the initial move you make to find the coefficient for \(x^{99}\) by summing 1 to 100.When the exercise progresses to deriving the coefficient of \(x^{98}\), the squared sum formula is used, which is: \((1+2+\ldots+100)^2 -(1^2 + 2^2 + \ldots + 100^2)\). This calculation, which avoids full expansion, is based on summing series squared and subtracting squares of the individual terms, making it suitable for larger seriestoo. These formulas simplify complex calculations into easier tasks, saving time and effort.