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 given by

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

we can follow these steps:

### Step 1: Identify the series and simplify
The series \(x + x^2 + x^3 + \ldots + x^n\) is a geometric series. The sum of a geometric series can be expressed as:

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

Thus, we can rewrite the limit as:

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

### Step 2: Substitute and simplify
Substituting \(x = 1\) directly leads to an indeterminate form \(0/0\). Therefore, we can apply L'Hôpital's Rule, which states that if we have a limit of the form \(0/0\), we can differentiate the numerator and denominator.

### Step 3: Differentiate the numerator and denominator
Differentiate the numerator:

1. The derivative of \(x + x^2 + x^3 + \ldots + x^n\) is:
\[ 1 + 2x + 3x^2 + \ldots + nx^{n-1}. \]

2. The derivative of \(-n\) is \(0\).

Thus, the derivative of the numerator becomes:
\[ 1 + 2x + 3x^2 + \ldots + nx^{n-1}. \]

The derivative of the denominator \(x - 1\) is \(1\).

### Step 4: Apply L'Hôpital's Rule
Now we can rewrite the limit using L'Hôpital's Rule:

\[ \lim_{x \to 1} (1 + 2x + 3x^2 + \ldots + nx^{n-1}). \]

### Step 5: Evaluate the limit
Substituting \(x = 1\):

\[ 1 + 2(1) + 3(1^2) + \ldots + n(1^{n-1}) = 1 + 2 + 3 + \ldots + n. \]

The sum \(1 + 2 + 3 + \ldots + n\) can be calculated using the formula:

\[ \frac{n(n + 1)}{2}. \]

### Step 6: Set the equation
Setting the above equal to \(820\):

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

Multiplying both sides by \(2\):

\[ n(n + 1) = 1640. \]

### Step 7: Rearrange into standard quadratic form
Rearranging gives:

\[ n^2 + n - 1640 = 0. \]

### Step 8: Solve the quadratic equation
Using the quadratic formula \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

Here, \(a = 1\), \(b = 1\), and \(c = -1640\):

\[ n = \frac{-1 \pm \sqrt{1 + 6560}}{2} = \frac{-1 \pm \sqrt{6561}}{2}. \]

Calculating \(\sqrt{6561} = 81\):

\[ n = \frac{-1 + 81}{2} = \frac{80}{2} = 40. \]

### Conclusion
Thus, the value of \(n\) is \(40\).