Calculate the sum of $n$ addends
$$7 + 77 + 777 +...+ 7... 7.$$
To calculate the sum of the series \( 7 + 77 + 777 + \ldots + \underbrace{777\ldots7}_{n \text{ digits}} \), we can use the hint provided and break down the problem step-by-step.

1. **Express each term in the series:**
   Each term in the series can be written as:
   \[
   7, \quad 77, \quad 777, \quad \ldots, \quad \underbrace{777\ldots7}_{n \text{ digits}}
   \]
   Notice that each term can be represented as \( 7 \times (1 + 10 + 10^2 + \ldots + 10^{k-1}) \) for \( k = 1, 2, \ldots, n \).

2. **Sum the series inside the parentheses:**
   The series inside the parentheses is a geometric series:
   \[
   1 + 10 + 10^2 + \ldots + 10^{k-1}
   \]
   The sum of the first \( k \) terms of a geometric series with the first term \( a = 1 \) and common ratio \( r = 10 \) is given by:
   \[
   S_k = \frac{10^k - 1}{10 - 1} = \frac{10^k - 1}{9}
   \]

3. **Sum the entire series:**
   Now, we need to sum the series:
   \[
   7 \left( \frac{10^1 - 1}{9} \right) + 7 \left( \frac{10^2 - 1}{9} \right) + 7 \left( \frac{10^3 - 1}{9} \right) + \ldots + 7 \left( \frac{10^n - 1}{9} \right)
   \]
   Factor out the common term \( 7 \):
   \[
   7 \left( \frac{(10^1 - 1) + (10^2 - 1) + (10^3 - 1) + \ldots + (10^n - 1)}{9} \right)
   \]

4. **Simplify the sum inside the parentheses:**
   \[
   (10^1 - 1) + (10^2 - 1) + (10^3 - 1) + \ldots + (10^n - 1)
   \]
   This can be written as:
   \[
   (10^1 + 10^2 + 10^3 + \ldots + 10^n) - n
   \]
   The sum of the geometric series \( 10^1 + 10^2 + 10^3 + \ldots + 10^n \) is:
   \[
   \frac{10(10^n - 1)}{10 - 1} = \frac{10^{n+1} - 10}{9}
   \]
   Therefore, the sum inside the parentheses becomes:
   \[
   \frac{10^{n+1} - 10}{9} - n
   \]

5. **Combine the results:**
   \[
   7 \left( \frac{\frac{10^{n+1} - 10}{9} - n}{9} \right) = 7 \left( \frac{10^{n+1} - 10 - 9n}{81} \right)
   \]

Thus, the final answer is:
\[
\boxed{7 \left( \frac{10^{n+1} - 9n - 10}{81} \right)}
\]