Wait, why do we reject the negative root here?

In the context of this problem, $n$ represents the number of terms in a polynomial, which must be a natural number. Since $n \in \mathbb{N}$, we discard the negative solution.

Why can't we just use energy conservation?

This is a calculus problem, not a physics problem! Energy conservation doesn't apply here. We use L'Hopital's Rule to resolve the indeterminate form.

Is there a faster way than L'Hopital?

You could factor the numerator by grouping, but L'Hopital's Rule is the most direct and elegant path for this specific limit.

JEE Main 2020 - 2 Sep (Morning)

LEVELJEE Main

Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $lim_{x \to 1} \frac{x + x ^{2} + x ^{3} + ... + x ^{n} - n}{x - 1} = 820$ , $(n \in N)$ then the value of $n$ is equal to

Enter Numerical Value:

Visualized Solution

Analyze the Limit Expression

Given limit: $lim_{x \to 1} \frac{x + x ^{2} + x ^{3} + ... + x ^{n} - n}{x - 1} = 820$
The variable $n$ is a natural number ( $n \in N$ ).
We need to find the value of $n$ that satisfies this equation.

Check the Form of the Limit

Substitute $x = 1$ into the expression to check the form.
Numerator: $1 + 1^{2} + 1^{3} + ... + 1^{n} - n = (1 + 1 + 1 + ... n times) - n = n - n = 0$ .
Denominator: $1 - 1 = 0$ .
The limit is in the indeterminate form $\frac{0}{0}$ .

Introduce L'Hopital's Rule

Since the form is $\frac{0}{0}$ , we apply L'Hopital's Rule.
L'Hopital's Rule: $lim_{x \to a} \frac{f ( x )}{g ( x )} = lim_{x \to a} \frac{f ^{'} ( x )}{g ^{'} ( x )}$ if the limit is $\frac{0}{0}$ or $\frac{\infty}{\infty}$ .

Differentiate Numerator and Denominator

Differentiate the numerator with respect to $x$ :
$\frac{d}{d x} (x + x^{2} + x^{3} + ... + x^{n} - n) = 1 + 2 x + 3 x^{2} + ... + n x^{n - 1} - 0$
Differentiate the denominator with respect to $x$ :
$\frac{d}{d x} (x - 1) = 1 - 0 = 1$

Apply the Limit $x \to 1$

Apply the limit $x \to 1$ to the new expression:
$lim_{x \to 1} \frac{1 + 2 x + 3 x ^{2} + ... + n x ^{n - 1}}{1} = 820$
Substitute $x = 1$ : $1 + 2 (1) + 3 (1)^{2} + ... + n (1)^{n - 1} = 820$
This simplifies to: $1 + 2 + 3 + ... + n = 820$

Sum of First $n$ Natural Numbers

Recall the formula for the sum of the first $n$ natural numbers:
$\sum_{k = 1}^{n} k = 1 + 2 + 3 + ... + n = \frac{n ( n + 1 )}{2}$
Substitute this into our equation: $\frac{n ( n + 1 )}{2} = 820$

Clear the Fraction

Multiply both sides by $2$ to eliminate the denominator:
$n (n + 1) = 820 \times 2$
$n (n + 1) = 1640$

Form the Quadratic Equation

Expand the left side and move all terms to one side:
$n^{2} + n = 1640$
$n^{2} + n - 1640 = 0$

Factorize the Quadratic

Factorize the quadratic equation by splitting the middle term:
$n^{2} + 41 n - 40 n - 1640 = 0$
$n (n + 41) - 40 (n + 41) = 0$
$(n - 40) (n + 41) = 0$

Solve for $n$

Set each factor to zero:
$n - 40 = 0 ⟹ n = 40$
$n + 41 = 0 ⟹ n = - 41$
Since $n \in N$ , we reject $n = - 41$ .
The only valid solution is $n = 40$ .

00:00 / 00:00

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Indeterminate Form

To solve the limit

lim_{x \to 1} \frac{x + x ^{2} + x ^{3} + ... + x ^{n} - n}{x - 1} = 820

, we must first evaluate the expression at the limit point

x = 1

Substituting

x = 1

into the numerator yields

1 + 1 + 1 + ... + 1 - n

, which simplifies to

n - n = 0

. The denominator similarly becomes

1 - 1 = 0

Since we have a

\frac{0}{0}

indeterminate form, we are justified in applying L'Hopital's Rule.

Applying L'Hopital's Rule

We differentiate the numerator and the denominator with respect to

x

independently.

The derivative of the numerator

x + x^{2} + x^{3} + ... + x^{n} - n

is:

1 + 2 x + 3 x^{2} + ... + n x^{n - 1}

The derivative of the denominator

x - 1

is simply

1

Now, we substitute

x = 1

into the resulting expression:

1 + 2 (1) + 3 (1)^{2} + ... + n (1)^{n - 1} = 820

Solving for n

The expression simplifies to the sum of the first

n

natural numbers:

1 + 2 + 3 + ... + n = 820

Using the standard formula for the sum of the first

n

integers, we have:

\frac{n ( n + 1 )}{2} = 820

Multiplying both sides by

2

gives the quadratic equation:

n (n + 1) = 1640

n^{2} + n - 1640 = 0

Factoring the quadratic equation, we find the roots are

n = 40

and

n = - 41

Since

n

must be a natural number, we discard the negative root. Therefore, the final answer is $n = 40$ .

Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $lim_{x \to 1} \frac{x + x ^{2} + x ^{3} + ... + x ^{n} - n}{x - 1} = 820$ , $(n \in N)$ then the value of $n$ is equal to

Enter Numerical Value:

Visualized Solution

Analyze the Limit Expression

Check the Form of the Limit

Introduce L'Hopital's Rule

Differentiate Numerator and Denominator

Apply the Limit $x \to 1$

Sum of First $n$ Natural Numbers

Clear the Fraction

Form the Quadratic Equation

Factorize the Quadratic

Solve for $n$

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Indeterminate Form

Applying L'Hopital's Rule

Solving for n

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Animated Solution for Mathematics - Limits, Continuity and Differentiability: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If limx→1​x−1x+x2+x3+...+xn−n​=820, (n∈N) then the value of n is equal to

Enter Numerical Value:

Visualized Solution

Analyze the Limit Expression

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Introduce L'Hopital's Rule

Differentiate Numerator and Denominator

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The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Indeterminate Form

Applying L'Hopital's Rule

Solving for n

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Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $lim_{x \to 1} \frac{x + x ^{2} + x ^{3} + ... + x ^{n} - n}{x - 1} = 820$ , $(n \in N)$ then the value of $n$ is equal to

Apply the Limit $x \to 1$

Sum of First $n$ Natural Numbers

Solve for $n$

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