Problem 22 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The coefficient of \(x^{49}\) in the product \((x-1)(x-3) \ldots\) \((x-99)\) is (A) \(-99^{2}\) (B) 1 (C) \(-2500\) (D) None of these

Step-by-Step Solution

Answer

The coefficient is -3725, so the answer is (D) None of these.

Step 1: Recognize the Problem Structure

This problem involves finding the coefficient of a particular power of \(x\) in the expansion of a polynomial product. The polynomial is a product of terms \((x-a_i)\) where \(a_i\) ranges from 1 to 99.

Step 2: Understand the Target Term

The term we want is \(x^{49}\). In the polynomial, the highest degree term is \(x^{99}\) (the degree of the polynomial). Thus, the term \(x^{49}\) is formed by choosing 50 constants from the terms \((x-1), (x-3), \ldots , (x-99)\).

Step 3: Choose the Coefficients

Using combinatorial selection, choose 50 factors from the 99 terms having \(-a_i\) as coefficients. This can be achieved by choosing 49 times \(-a_i\), and leaving out 50 \(x-a_i\) terms.

Step 4: Calculate Sign

Each term \((x- a_i)\) contributes a sign change depending on whether \(x\) or \(-a_i\) is chosen. The target term \(x^{49}\) emerges with choosing 49 \(-a_i\)'s, leading to an overall negative sign (as 49 is odd). Therefore, the coefficient is negative.

Step 5: Combine to Form Equation

Coefficients' result becomes the sum of choosing product constants, which are the combination of terms excluded, specifically\(-\sum a_i\) where \(i\) refers to the indices of omitted terms (50 terms).The summation from 1 to 99 must exclude specific piecewise sums, but realizing it ends similarly. Direct computation gives \[-\sum_{i=50}^{99} i = -\sum_{i=1}^{99} i + \sum_{i=1}^{49} i \].

Step 6: Final Calculation

Direct calculation of sums yields: \[\sum_{i=1}^{99} i = \frac{99 \times 100}{2} = 4950\] and \[\sum_{i=1}^{49} i = \frac{49 \times 50}{2} = 1225\]. Thus:\[-4950 + 1225 = -3725\].No provided option matches this, indicating possible inclusion in option (D).

Key Concepts

Product of Linear Factors

The concept of the product of linear factors is central to polynomial expansions in algebra. Linear factors are expressions of the form
- \((x-a)\), where each factor primarily contributes one root to the polynomial. When these factors are multiplied, they form a polynomial of a degree equal to the number (
) of factors.
For example, if we have a product like \((x-1)(x-2)...(x-n)\), this creates a polynomial of degree \(n\) since there are \(n\) factors. The highest degree term, in this case, would
be \(x^n\).
It's important to understand that every additional linear factor increases the degree of the polynomial by one (
).
This exercise involves a polynomial given by the product of 99 such linear factors, reaching a degree of 99, and we are interested in determining specific coefficients within its expansion.

Combinatorial Selection

Combinatorial selection involves choosing a subset of elements from a larger set (
) without regard to order. In the context of polynomial expansions, particularly with products of linear factors, this
becomes essential in determining the coefficients of terms of specific degrees.
In our exercise, we have linear factors from \((x-1)\) to \((x-99)\). To find the coefficient of \(x^{49}\), we need to select exactly 50 constant terms (the \((-a_i)\)). This is done by excluding 50 terms, achieved by combinatorially choosing which of the factors should contribute a constant (
) instead of the variable \(x\).
Once this selection is imagined, the coefficient is formed by the sum of the constants of the omitted (
) factors, highlighting a key relationship between selection and summation in polynomial expansion.

Polynomial Expansion

Polynomial expansion translates the product of multiple linear factors into an expanded algebraic expression (
). In this expanded form, you'll find monomials of all possible degrees from zero to the highest degree determined by the number of multiplied factors.
The trick is to correctly compute the coefficient of each term, especially the intermediate ones like \(x^{49}\) in our involved polynomial of degree \(99\).
Finding coefficients relies on understanding the structure of the polynomial (
) and the process of expansion. Each combination of the factors that results in a particular power of \(x\) contributes to
that term's coefficient.
We typically use the sum of constants from selected factors, influenced by understanding the pattern of expansion, as seen in our current polynomial
example, where an algebraic approach leads to calculating a precise numerical representation of desired coefficients.