Problem 37 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If \(\lim _{x \rightarrow 1} \frac{x+x^{2}+x^{3}+\ldots+x^{n}-n}{x-1}=820,(n \in \mathbf{N})\) then the value of \(n\) is equal to

Step-by-Step Solution

Answer

The value of \( n \) is 821.

Step 1: Understand the Limit Expression

The expression we are looking at is \( \lim _{x \rightarrow 1} \frac{x+x^{2} + x^{3} + \ldots + x^{n} - n}{x-1} \). This limit is being evaluated as \( x \) approaches 1, which suggests we need to handle any indeterminate forms possibly by using L'Hôpital's Rule or expanding terms.

Step 2: Recognize the Sum as a Series

The numerator \( x + x^2 + \ldots + x^n \) is a geometric series. The sum of this series is \( \frac{x(x^n - 1)}{x - 1} \). Subtracting \( n \), the numerator becomes \( \frac{x(x^n - 1)}{x - 1} - n \).

Step 3: Simplify the Expression

We substitute the geometric series sum into the expression: \( \frac{\frac{x(x^n - 1)}{x - 1} - n}{x-1} \). To simplify further, multiply through by \( (x-1) \) to clear the fraction in the numerator, resulting in: \( x(x^n-1) - n(x-1) \).

Step 4: Evaluate the Limit Using Derivatives

Since the expression is in an indeterminate form \( \frac{0}{0} \) as \( x \rightarrow 1 \), apply L'Hôpital's Rule by differentiating the numerator and the denominator. The numerator \( x(x^n - 1) - n(x-1) \) becomes \( nx^n - 1 \) after differentiating, and the denominator becomes 1.

Step 5: Solve for n

Evaluate the differentiated limit: \( \lim_{x \to 1} (nx^n - 1) = n - 1 \). Set this equal to \( 820 \) (since we have \( \lim_{x \to 1} \cdots = 820 \)), giving us \( n - 1 = 820 \) so \( n = 821 \).

Key Concepts

Geometric Series

A geometric series is a sum of terms where each term is the product of the previous term and a constant, known as the common ratio. In our context, the numerator of the expression is a geometric series: \( x + x^2 + x^3 + \ldots + x^n \). The formula to sum a geometric series where the first term is \( a \) and the common ratio is \( r \) is given by:

• \( S_n = a \frac{r^n - 1}{r - 1} \)

For our problem, the first term \( a \) is \( x \) and the common ratio \( r \) is also \( x \), yielding the sum as \( \frac{x(x^n - 1)}{x - 1} \). This formula simplifies how we handle the series in calculation, particularly under a limit, as it enables us to manage expressions cleanly.
Subtracting \( n \) from this series leads to the adjustment necessary for evaluating the limit in the limit expression. Recognizing these series and using their sum formulas is crucial in calculus as it allows complex expressions to be simplified down to more manageable terms.

L'Hôpital's Rule

L'Hôpital's Rule is a valuable tool in calculus for dealing with indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). When an expression results in such a form upon directly substituting the limit, it indicates we need further analysis.
The rule states that if \( \lim_{x \to c} \frac{f(x)}{g(x)} \) results in an indeterminate form, we can find the limit by differentiating the numerator \( f(x) \) and the denominator \( g(x) \), and then calculating \( \lim_{x \to c} \frac{f'(x)}{g'(x)} \) instead.

• This is exactly what we apply once we rewrite and simplify our limit expression in this exercise.
• We differentiate \( x(x^n - 1) - n(x - 1) \) with respect to \( x \), producing \( nx^n - 1 \).

This process proves the utility of L'Hôpital's Rule in simplifying complex limits, especially when direct substitution doesn't work due to zero in both the numerator and denominator.

Indeterminate Forms

Indeterminate forms occur in limits when substitution leads to ambiguous or undefined expressions, such as \( \frac{0}{0} \) or \( \infty - \infty \). These forms require careful analysis, often involving algebraic manipulation or calculus techniques like L'Hôpital's Rule.

• In our problem, as \( x \to 1 \), both the numerator and denominator zero out.
• This situation indicates a \( \frac{0}{0} \) indeterminate form, which signals us to potentially apply L'Hôpital’s Rule.

By transforming the original expression using geometric series and then applying L'Hôpital’s Rule, we see the power of converting an indeterminate form into something we can evaluate. A strong understanding of these forms and how to resolve them is fundamental in calculus, allowing us to solve seemingly unsolvable limits effectively.