Find the sum of the series: `1. n+2.(n-1)+3.(n-2)++(n-1). 2+n .1.`

Explanation

To find the sum of the series \( S_n = 1 \cdot n + 2 \cdot (n-1) + 3 \cdot (n-2) + \ldots + (n-1) \cdot 2 + n \cdot 1 \), we can break down the solution step by step.

### Step 1: Identify the General Term
The \( r \)-th term of the series can be expressed as:
\[ T_r = r \cdot (n - r + 1) \]
This is because the first part \( r \) goes from 1 to \( n \) and the second part \( (n - r + 1) \) decreases from \( n \) to 1.

### Step 2: Write the Sum
Now, we can express the sum \( S_n \) as:
\[ S_n = \sum_{r=1}^{n} T_r = \sum_{r=1}^{n} r \cdot (n - r + 1) \]

### Step 3: Expand the Sum
Expanding the term gives:
\[ S_n = \sum_{r=1}^{n} (r \cdot n - r^2 + r) \]
This can be separated into three different sums:
\[ S_n = n \sum_{r=1}^{n} r - \sum_{r=1}^{n} r^2 + \sum_{r=1}^{n} r \]

### Step 4: Use Summation Formulas
We can use the formulas for the sums of the first \( n \) natural numbers and the sum of the squares of the first \( n \) natural numbers:
1. The sum of the first \( n \) natural numbers:
\[ \sum_{r=1}^{n} r = \frac{n(n+1)}{2} \]
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 5: Substitute the Formulas
Substituting these formulas into our expression for \( S_n \):
\[ S_n = n \cdot \frac{n(n+1)}{2} - \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \]

### Step 6: Combine Like Terms
Now, we can combine the terms:
\[ S_n = \frac{n(n+1)}{2} \left( n + 1 - \frac{(2n+1)}{3} \right) \]
To simplify, we need a common denominator:
\[ S_n = \frac{n(n+1)}{2} \left( \frac{3(n+1) - (2n + 1)}{3} \right) \]
This simplifies to:
\[ S_n = \frac{n(n+1)}{2} \cdot \frac{n + 2}{3} \]

### Step 7: Final Result
Thus, the final result for the sum of the series is:
\[ S_n = \frac{n(n+1)(n+2)}{6} \]