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Animated Solution for Mathematics - Quadratic Equations: The coefficient of $x^{99}$ in the polynomial $(x - 1) (x - 2) \dots (x - 100)$ is .........

Q: Wait, why is the coefficient negative? Shouldn't it be positive?

Great question! Remember the form $P(x) = (x-r_1)(x-r_2)\dots(x-r_n)$. When you multiply these, the $x^{n-1}$ term comes from picking $x$ from $n-1$ brackets and the constant $-r_i$ from the $i$-th bracket. Since every constant is negative, the sum of these constants will be negative. It is a direct consequence of the structure of the factors.

Q: Can I use the Binomial Theorem here?

Not directly. The Binomial Theorem applies to $(x+a)^n$, where the constant is the same in every bracket. Here, the constants are different ($-1, -2, \dots, -100$), so we must use the fundamental principle of polynomial expansion or Vieta's relations.

Enter Numerical Value:

Visualized Solution

Visualizing the Polynomial $P (x)$

We are given the polynomial $P (x) = (x - 1) (x - 2) (x - 3) \dots (x - 100)$ .
This is a product of $100$ linear factors.
Therefore, the degree of this polynomial is exactly $100$ .

How Terms are Formed in Expansion

When we multiply $n$ brackets, each term in the expansion is formed by choosing either $x$ or the constant from each bracket.
To get the term $x^{100}$ , we must choose $x$ from all $100$ brackets.
There is only $1$ way to do this, giving us $1 \cdot x^{100}$ .

Forming the $x^{99}$ Term

To get a term with $x^{99}$ , we must choose $x$ from exactly $99$ brackets.
From the remaining $1$ bracket, we must choose the constant term.
For example, choosing constants $- 1$ , $- 2$ , or $- 100$ while choosing $x$ from all other brackets.

Summing the Contributions

Let's write down the sum of all these individual terms:
Term $= (- 1) x^{99} + (- 2) x^{99} + (- 3) x^{99} + \dots + (- 100) x^{99}$
Factoring out $x^{99}$ , we get: $- (1 + 2 + 3 + \dots + 100) x^{99}$

Identifying the Coefficient

The coefficient of $x^{99}$ is therefore: $- (1 + 2 + 3 + \dots + 100)$
This is the negative sum of the first $100$ positive integers.
This is also directly related to Vieta's relations, where the coefficient of $x^{n - 1}$ is $- \sum roots$ .

Recalling the Sum of Arithmetic Progression

To find the sum of the first $n$ natural numbers, we use the formula:
$S_{n} = \frac{n ( n + 1 )}{2}$
This is the formula for the sum of an Arithmetic Progression (AP).

Substituting $n = 100$

Here, we need the sum of the first $100$ natural numbers, so we set $n = 100$ .
Substituting $n = 100$ into the formula:
Sum $= \frac{100 ( 100 + 1 )}{2}$

Simplifying the Fraction

Simplify the term inside the parenthesis: $100 + 1 = 101$
The expression becomes: $\frac{100 \times 101}{2}$
Now, divide $100$ by $2$ to get $50$ .

Computing the Product

Multiply $50$ by $101$ :
$50 \times 101 = 5050$
So, the sum of the first $100$ natural numbers is $5050$ .

Applying the Negative Sign

Recall that the coefficient of $x^{99}$ is the negative of this sum.
Coefficient of $x^{99} = - 5050$
This is our final answer!

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The Sigma Insight: Relation Between Roots and Coefficients

Solution Diagram

Analyzing the Setup

Imagine standing before a massive wall of 100 brackets, all lined up:

P (x) = (x - 1) (x - 2) (x - 3) \dots (x - 100)

. This is a polynomial of degree 100 because each of the 100 brackets contributes exactly one

x

to the product.

If you were to multiply them all out, the highest power you could possibly form is

x^{100}

. Today, we are hunting for the coefficient of

x^{99}

The Combinatorial Logic

To understand how the

x^{99}

term is born, we must think like a combinatorialist. When you expand a product of brackets, you are essentially making a choice for every single bracket.

From each bracket, you must pick either the

x

or the constant term. To end up with

x^{99}

, you must choose the

x

from exactly 99 brackets. This leaves you with exactly one bracket from which you must choose the constant term.

There are 100 brackets. You can choose the constant from the first bracket (

- 1

), or the second (

- 2

), or the third (

- 3

), and so on, all the way to the 100th (

- 100

). Each of these choices creates a term of the form

- k \cdot x^{99}

The Summation

If we sum all these individual contributions, we get:

(- 1) x^{99} + (- 2) x^{99} + (- 3) x^{99} + \dots + (- 100) x^{99}

By factoring out the

x^{99}

, we are left with the negative sum of the first 100 natural numbers:

- (1 + 2 + 3 + \dots + 100) x^{99}

This is where the beauty of Vieta's relations shines. Vieta's formulas tell us that for any monic polynomial, the coefficient of the second-highest power is always the negative sum of the roots. Here, our roots are

1, 2, 3, \dots, 100

The Final Calculation

Now, we calculate the sum of the first 100 integers using the elegant formula for the sum of an arithmetic progression:

S_{n} = \frac{n ( n + 1 )}{2}

Substituting

n = 100

, we get:

\frac{100 ( 100 + 1 )}{2} = \frac{100 \times 101}{2} = 50 \times 101 = 5050

Since our coefficient is the negative of this sum, the final coefficient of

x^{99}

is $- 5050$ . This is a clean, satisfying result that emerges from the deep symmetry of the polynomial.

Common Traps & FAQs

Wait, why is the coefficient negative? Shouldn't it be positive?

Great question! Remember the form

P (x) = (x - r_{1}) (x - r_{2}) \dots (x - r_{n})

. When you multiply these, the

x^{n - 1}

term comes from picking

x

from

n - 1

brackets and the constant

- r_{i}

from the

i

-th bracket. Since every constant is negative, the sum of these constants will be negative. It is a direct consequence of the structure of the factors.

Can I use the Binomial Theorem here?

Not directly. The Binomial Theorem applies to

(x + a)^{n}

, where the constant is the same in every bracket. Here, the constants are different (

- 1, - 2, \dots, - 100

), so we must use the fundamental principle of polynomial expansion or Vieta's relations.

Is there a faster way than writing out the sum?

Yes! That is the beauty of Vieta's relations. Instead of expanding, recognize that the coefficient of

x^{n - 1}

is simply the negative sum of the roots. Once you identify the roots as

1, 2, \dots, 100

, the problem reduces to a simple arithmetic series sum, which is much faster.

Animated Solution for Mathematics - Quadratic Equations: The coefficient of $x^{99}$ in the polynomial $(x - 1) (x - 2) \dots (x - 100)$ is .........

Enter Numerical Value:

Visualized Solution

Visualizing the Polynomial $P (x)$

How Terms are Formed in Expansion

Forming the $x^{99}$ Term

Summing the Contributions

Identifying the Coefficient

Recalling the Sum of Arithmetic Progression

Substituting $n = 100$

Simplifying the Fraction

Computing the Product

Applying the Negative Sign

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

The Combinatorial Logic

The Summation

The Final Calculation

Similar Questions

Let $α, β$ be the roots of the equation $x^{2} - 4 λ x + 5 = 0$ and $α, γ$ be the roots of the equation $x^{2} - (32 + 23) x + 7 + 3 λ 3 = 0$ . If $β + γ = 32$ , then $(α + 2 β + γ)^{2}$ is equal to ______.

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Animated Solution for Mathematics - Quadratic Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The coefficient of x99 in the polynomial (x−1)(x−2)…(x−100) is .........

Enter Numerical Value:

Visualized Solution

Visualizing the Polynomial .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(x)

How Terms are Formed in Expansion

Forming the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x99 Term

Summing the Contributions

Identifying the Coefficient

Recalling the Sum of Arithmetic Progression

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n=100

Simplifying the Fraction

Computing the Product

Applying the Negative Sign

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

The Combinatorial Logic

The Summation

The Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let α and β be two real roots of the (k+1)tan2x−2​⋅λtanx=(1−k), where k(=−1) and λ are real numbers. If tan2(α+β)=50, then a value of λ is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let α and β be the roots of the equation x2−x−1=0. If pk​=(α)k+(β)k,k≥1, then which of the following statements is not true?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let α and β are the roots of the equation x2−x−1=0. If pk​=(α)k+(β)k,k≥1 then which one of the following statements is not true?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x) be a quadratic polynomial such that f(−2)+f(3)=0. If one of the roots of f(x)=0 is −1, then the sum of the roots of f(x)=0 is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If (x2+px+1) is a factor of (ax3+bx+c), then

Animated Solution for Mathematics - Quadratic Equations: The coefficient of $x^{99}$ in the polynomial $(x - 1) (x - 2) \dots (x - 100)$ is .........

Visualizing the Polynomial $P (x)$

Forming the $x^{99}$ Term

Substituting $n = 100$

Let $α$ and $β$ be two real roots of the $(k + 1) tan^{2} x - 2 \cdot λ tan x = (1 - k)$ , where $k (\neq = - 1)$ and $λ$ are real numbers. If $tan^{2} (α + β) = 50$ , then a value of $λ$ is:

Let $α$ and $β$ be the roots of the equation $x^{2} - x - 1 = 0$ . If $p_{k} = (α)^{k} + (β)^{k}, k \geq 1$ , then which of the following statements is not true?

Let $α$ and $β$ are the roots of the equation $x^{2} - x - 1 = 0$ . If $p_{k} = (α)^{k} + (β)^{k}, k \geq 1$ then which one of the following statements is not true?

Let $f (x)$ be a quadratic polynomial such that $f (- 2) + f (3) = 0$ . If one of the roots of $f (x) = 0$ is $- 1$ , then the sum of the roots of $f (x) = 0$ is equal to:

If $(x^{2} + p x + 1)$ is a factor of $(a x^{3} + b x + c)$ , then