Sum up `1 + (1+2)+(1+ 2+3) +...+(1+2+3+...+ n )`.

Answer

To solve the problem of summing up the series \( S = 1 + (1 + 2) + (1 + 2 + 3) + \ldots + (1 + 2 + 3 + \ldots + n) \), we can follow these steps:

### Step 1: Identify the nth term
The nth term of the series can be expressed as:
\[ T_r = 1 + 2 + 3 + \ldots + r \]
This is the sum of the first \( r \) natural numbers.

### Step 2: Use the formula for the sum of the first r natural numbers
The formula for the sum of the first \( r \) natural numbers is:
\[ T_r = \frac{r(r + 1)}{2} \]

### Step 3: Write the total sum S
Now, we can express the total sum \( S \) as:
\[ S = \sum_{r=1}^{n} T_r = \sum_{r=1}^{n} \frac{r(r + 1)}{2} \]

### Step 4: Factor out the constant
We can factor out \( \frac{1}{2} \) from the sum:
\[ S = \frac{1}{2} \sum_{r=1}^{n} r(r + 1) \]

### Step 5: Expand the summation
Now, we can expand \( r(r + 1) \):
\[ S = \frac{1}{2} \sum_{r=1}^{n} (r^2 + r) = \frac{1}{2} \left( \sum_{r=1}^{n} r^2 + \sum_{r=1}^{n} r \right) \]

### Step 6: Use the formulas for the sums
We know the formulas for these sums:
- The sum of the first \( n \) natural numbers:
\[ \sum_{r=1}^{n} r = \frac{n(n + 1)}{2} \]
- The sum of the squares of the first \( n \) natural numbers:
\[ \sum_{r=1}^{n} r^2 = \frac{n(n + 1)(2n + 1)}{6} \]

### Step 7: Substitute the formulas into S
Substituting these formulas into our expression for \( S \):
\[ S = \frac{1}{2} \left( \frac{n(n + 1)(2n + 1)}{6} + \frac{n(n + 1)}{2} \right) \]

### Step 8: Simplify the expression
Now we can simplify:
\[ S = \frac{1}{2} \left( \frac{n(n + 1)(2n + 1)}{6} + \frac{3n(n + 1)}{6} \right) \]
\[ = \frac{1}{2} \cdot \frac{n(n + 1)(2n + 1 + 3)}{6} \]
\[ = \frac{1}{2} \cdot \frac{n(n + 1)(2n + 4)}{6} \]
\[ = \frac{n(n + 1)(n + 2)}{12} \]

### Final Result
Thus, the sum \( S \) is:
\[ S = \frac{n(n + 1)(n + 2)}{6} \]