If `lim_(x rarr 1)((x+x^2+x^3+....+x^n-n)/(x-1))=820`, then find n.

Answer

To solve the limit problem, we need to evaluate the limit:

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

### Step 1: Simplify the series

The series \(x + x^2 + x^3 + \ldots + x^n\) can be expressed as a geometric series. The sum of the first \(n\) terms of a geometric series is given by:

\[ S_n = \frac{x(1 - x^n)}{1 - x} \]

Thus, we can rewrite our limit as:

\[ \lim_{x \to 1} \frac{\frac{x(1 - x^n)}{1 - x} - n}{x - 1} \]

### Step 2: Combine the fractions

We can combine the terms in the numerator:

\[ \frac{x(1 - x^n) - n(1 - x)}{(1 - x)(x - 1)} = \frac{x - x^{n+1} - n + nx}{(1 - x)(x - 1)} \]

This simplifies to:

\[ \frac{(n + 1)x - x^{n+1} - n}{(1 - x)(x - 1)} \]

### Step 3: Apply L'Hôpital's Rule

Since both the numerator and denominator approach 0 as \(x\) approaches 1, we can apply L'Hôpital's Rule. We differentiate the numerator and the denominator:

- The derivative of the numerator \( (n + 1)x - x^{n + 1} - n \) is \( (n + 1) - (n + 1)x^n \).
- The derivative of the denominator \( (1 - x)(x - 1) \) is \(-1\).

Thus, we have:

\[ \lim_{x \to 1} \frac{(n + 1) - (n + 1)x^n}{-1} \]

### Step 4: Evaluate the limit

Substituting \(x = 1\):

\[ \lim_{x \to 1} \frac{(n + 1) - (n + 1) \cdot 1^n}{-1} = \frac{(n + 1) - (n + 1)}{-1} = \frac{0}{-1} = 0 \]

This means we need to differentiate again since we still have an indeterminate form.

### Step 5: Differentiate again

Differentiate the numerator again:

The derivative of \( (n + 1) - (n + 1)x^n \) is:

\[ -(n + 1)nx^{n-1} \]

And the derivative of the denominator remains \(-1\).

Thus, we have:

\[ \lim_{x \to 1} \frac{-(n + 1)nx^{n-1}}{-1} = (n + 1)n \]

### Step 6: Set the limit equal to 820

Setting this equal to 820:

\[ (n + 1)n = 820 \]

### Step 7: Rearrange the equation

This can be rearranged to:

\[ n^2 + n - 820 = 0 \]

### Step 8: Solve the quadratic equation

Using the quadratic formula:

\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

where \(a = 1\), \(b = 1\), and \(c = -820\):

\[ n = \frac{-1 \pm \sqrt{1 + 3280}}{2} = \frac{-1 \pm \sqrt{3281}}{2} \]

Calculating the square root:

\[ \sqrt{3281} \approx 57.25 \]

Thus, we have:

\[ n = \frac{-1 + 57.25}{2} \approx 28.125 \quad \text{(not an integer)} \]
or
\[ n = \frac{-1 - 57.25}{2} \quad \text{(not valid)} \]

### Step 9: Check for integer solutions

We can also factor \(n^2 + n - 820\):

\[ (n - 40)(n + 41) = 0 \]

Thus, \(n = 40\) or \(n = -41\). Since \(n\) must be a positive integer, we conclude:

\[ n = 40 \]

### Final Answer

The value of \(n\) is:

\[ \boxed{40} \]