Key Concepts and Formulas

• L'Hôpital's Rule: If a limit of a function results in an indeterminate form like $\frac{0}{0}$ or $\frac{\infty}{\infty}$, L'Hôpital's Rule can be applied. It states that the limit of the ratio of two functions is equal to the limit of the ratio of their derivatives, provided the latter limit exists. Mathematically, if $\mathop {\lim }\limits_{x \to c} f(x) = 0$ and $\mathop {\lim }\limits_{x \to c} g(x) = 0$, then $\mathop {\lim }\limits_{x \to c} {{f(x)} \over {g(x)}} = \mathop {\lim }\limits_{x \to c} {{f'(x)} \over {g'(x)}}$.
• Sum of the first n natural numbers: The sum of the first $n$ natural numbers is given by the formula $S_n = 1 + 2 + 3 + ... + n = {{n(n+1)} \over 2}$.
• Geometric Series Sum: The sum of a finite geometric series is given by $a + ar + ar^2 + ... + ar^{n-1} = a{{r^n - 1} \over {r - 1}}$. While not directly used in the most efficient solution, understanding the terms $x, x^2, ..., x^n$ as a geometric series is relevant to the numerator's structure.

Step-by-Step Solution

Step 1: Analyze the given limit and identify the indeterminate form. We are given the limit: $L = \mathop {\lim }\limits_{x \to 1} {{x + {x^2} + {x^3} + ... + {x^n} - n} \over {x - 1}}$ As $x \to 1$, the numerator approaches $1 + 1^2 + 1^3 + ... + 1^n - n$. This sum is $1 + 1 + 1 + ... + 1$ ($n$ times) $- n$, which equals $n - n = 0$. The denominator approaches $1 - 1 = 0$. Since the limit results in the indeterminate form $\frac{0}{0}$, we can apply L'Hôpital's Rule.

Step 2: Apply L'Hôpital's Rule. We need to find the derivative of the numerator and the derivative of the denominator with respect to $x$. Let $f(x) = x + x^2 + x^3 + ... + x^n - n$. Let $g(x) = x - 1$.

The derivative of the numerator is: $f'(x) = \frac{d}{dx}(x + x^2 + x^3 + ... + x^n - n)$ $f'(x) = \frac{d}{dx}(x) + \frac{d}{dx}(x^2) + \frac{d}{dx}(x^3) + ... + \frac{d}{dx}(x^n) - \frac{d}{dx}(n)$ Using the power rule, $\frac{d}{dx}(x^k) = kx^{k-1}$: $f'(x) = 1 + 2x + 3x^2 + ... + nx^{n-1} - 0$ $f'(x) = 1 + 2x + 3x^2 + ... + nx^{n-1}$

The derivative of the denominator is: $g'(x) = \frac{d}{dx}(x - 1)$ $g'(x) = 1 - 0$ $g'(x) = 1$

Now, apply L'Hôpital's Rule: $L = \mathop {\lim }\limits_{x \to 1} {{f'(x)} \over {g'(x)}} = \mathop {\lim }\limits_{x \to 1} {{1 + 2x + 3x^2 + ... + nx^{n-1}} \over 1}$

Step 3: Evaluate the limit after applying L'Hôpital's Rule. Substitute $x = 1$ into the expression obtained in Step 2: $L = 1 + 2(1) + 3(1)^2 + ... + n(1)^{n-1}$ $L = 1 + 2 + 3 + ... + n$

Step 4: Use the formula for the sum of the first n natural numbers. The sum $1 + 2 + 3 + ... + n$ is the sum of the first $n$ natural numbers, which is given by the formula: $S_n = {{n(n+1)} \over 2}$ So, the limit is equal to: $L = {{n(n+1)} \over 2}$

Step 5: Set the limit equal to the given value and solve for n. We are given that the limit is equal to 820: $L = 820$ Therefore, ${{n(n+1)} \over 2} = 820$

Step 6: Solve the quadratic equation for n. Multiply both sides by 2: $n(n+1) = 1640$ Expand the left side: $n^2 + n = 1640$ Rearrange into a standard quadratic equation: $n^2 + n - 1640 = 0$ We need to find two numbers that multiply to -1640 and add to 1. We can factor this quadratic or use the quadratic formula. Let's try factoring by looking for factors of 1640 that are close to each other. $1640 = 164 \times 10 = 4 \times 41 \times 2 \times 5 = 8 \times 41 \times 5 = 40 \times 41$. The factors 40 and 41 have a difference of 1. So, we can set up the factorization as: $(n + 41)(n - 40) = 0$ This gives two possible solutions for $n$: $n + 41 = 0 \implies n = -41$ $n - 40 = 0 \implies n = 40$

Step 7: Select the valid value of n. The problem states that $n \in \mathbb{N}$ (n is a natural number). Natural numbers are positive integers (1, 2, 3, ...). Therefore, $n = -41$ is not a valid solution. The valid solution is $n = 40$.

Common Mistakes & Tips

• Incorrectly applying L'Hôpital's Rule: Ensure the limit is indeed in an indeterminate form ($\frac{0}{0}$ or $\frac{\infty}{\infty}$) before applying the rule. Incorrectly differentiating the terms in the numerator is a common error.
• Algebraic Errors in Factoring/Solving Quadratic Equations: Double-check your calculations when solving the quadratic equation $n^2 + n - 1640 = 0$. Mistakes in factoring or using the quadratic formula can lead to the wrong value of $n$.
• Forgetting the Domain of n: The problem specifies that $n \in \mathbb{N}$. Always ensure your final answer for $n$ is a natural number.

Summary

The given limit evaluates to an indeterminate form of $\frac{0}{0}$, which allows us to apply L'Hôpital's Rule. After differentiating the numerator and the denominator, we find that the limit simplifies to the sum of the first $n$ natural numbers, $1 + 2 + ... + n$. Using the formula for this sum, $\frac{n(n+1)}{2}$, we set it equal to the given value of 820. Solving the resulting quadratic equation $n^2 + n - 1640 = 0$ yields two solutions, $n = 40$ and $n = -41$. Since $n$ must be a natural number, we select $n = 40$.

The final answer is \boxed{40}.