Problem 1
Question
The coefficient of \(x^{17}\) in the expansion of \((x-1)(x-2)(x-3) \ldots(x-18)\) is (A) \(\frac{171}{2}\) (B) 342 (C) \(-171\) (D) 684
Step-by-Step Solution
Verified Answer
The coefficient of \(x^{17}\) is -171.
1Step 1: Understand the Problem
We need to find the coefficient of \(x^{17}\) in the expansion of the product \((x-1)(x-2)(x-3) \ldots (x-18)\). This is equivalent to finding the sum of all the coefficients of the first-degree terms when we expand and combine the product.
2Step 2: Apply Viète's Formula
The expression \((x-1)(x-2)(x-3) \ldots (x-18)\) is a polynomial of degree 18. By Viète's formulas, the sum of the roots of the equation set to 0, which are 1, 2, 3, ..., 18, gives the negative coefficient of the \(x^{17}\) term.
3Step 3: Calculate the Sum of Roots
Find the sum of the integers from 1 to 18 using the formula for the sum of an arithmetic series: \(S = \frac{n}{2}(a + l)\), where \(n\) is the number of terms, \(a\) is the first term, and \(l\) is the last term. Here, \(S = \frac{18}{2}(1 + 18) = 9 \times 19 = 171\).
4Step 4: Determine the Coefficient
The coefficient of \(x^{17}\) is the negative of the sum of the roots, which we calculated as 171. Therefore, the coefficient of \(x^{17}\) is \(-171\).
Key Concepts
Viète's FormulasPolynomial RootsArithmetic Series
Viète's Formulas
Viète's formulas are a powerful tool in algebra, specifically when dealing with polynomials and their roots. They're a handy way to connect the coefficients of a polynomial to its roots. Viète's formulas state that for a polynomial equation of the form \(a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 = 0\):
- The sum of the roots \(r_1 + r_2 + \cdots + r_n\) is equal to \(-\frac{a_{n-1}}{a_n}\).
- The product of the roots \(r_1 r_2 \cdots r_n\) for an even number of roots is \(\frac{a_0}{a_n}\).
- For an odd number of roots, the product of the roots is \(-\frac{a_0}{a_n}\).
Polynomial Roots
Polynomial roots are the values of \(x\) that make a polynomial equation equal to zero. These roots are the foundation in the context of polynomial factorization and expansion. For the polynomial \(f(x) = (x-1)(x-2)\cdots(x-18)\), the roots are simply the integers 1 through 18.Finding these roots allows us to understand more about the polynomial's behavior and structure. The coefficients of the polynomial in standard expanded form relate to these roots as described by Viète's formulas. For example:
- The roots determine how terms combine to form specific powers of \(x\) in the expansion.
- Each root contributes to the terms in a way that defines the shape of the polynomial graph.
Arithmetic Series
An arithmetic series is a sequence of numbers in which the difference of any two successive members is a constant. This is remnant in the formula for finding the sum of such a series. In this task, it assists in finding the sum of the roots of the polynomial \((x-1)(x-2)\cdots(x-18)\).To find the sum of the integers from 1 to 18, which constitute an arithmetic series, we use the arithmetic series sum formula: \[S = \frac{n}{2}(a + l)\]
- \(n\) is the number of terms.
- \(a\) is the first term in the series.
- \(l\) is the last term.
- \(S = \frac{18}{2}(1 + 18) = 9 \times 19= 171\).
Other exercises in this chapter
Problem 2
The fractional part of \(\frac{2^{4 n}}{15}\) is (A) \(\frac{2}{15}\) (B) \(\frac{1}{15}\) (C) \(\frac{4}{15}\) (D) none of these
View solution Problem 4
The sum of the coefficients of all the integral powers of \(\mathrm{x}\) in the expansion of \((1+2 \sqrt{x})^{80}\) is (A) \(\frac{1}{2}\left(3^{80}+1\right)\)
View solution Problem 5
If \([x]\) denotes the greatest integer less than or equal to \(x\), then \(\left[(6 \sqrt{6}+14)^{2 n+1}\right]\) (A) is an even integer (B) is an odd integer
View solution