Sum of the series ` n.1 + (n - 1).2 + (n - 2).3 + …….1` is……………….

Answer

To find the sum of the series \( n \cdot 1 + (n - 1) \cdot 2 + (n - 2) \cdot 3 + \ldots + 1 \), we can follow these steps:

### Step 1: Identify the general term of the series
The \( r \)-th term of the series can be expressed as:
\[ T_r = (n - r + 1) \cdot r \]
where \( r \) ranges from 1 to \( n \).

### Step 2: Expand the general term
Expanding the general term gives:
\[ T_r = n \cdot r - r^2 + r \]
This can be rewritten as:
\[ T_r = n \cdot r - r^2 + r = n \cdot r - r^2 + r \]

### Step 3: Write the sum of the series
The sum of the series \( S \) can be expressed as:
\[ S = \sum_{r=1}^{n} T_r = \sum_{r=1}^{n} (n \cdot r - r^2 + r) \]

### Step 4: Separate the summation
We can separate the summation into three parts:
\[ S = n \sum_{r=1}^{n} r - \sum_{r=1}^{n} r^2 + \sum_{r=1}^{n} r \]

### Step 5: Use formulas for summation
Using the formulas for the sum of the first \( n \) natural numbers and the sum of the squares of the first \( n \) natural numbers:
1. \( \sum_{r=1}^{n} r = \frac{n(n + 1)}{2} \)
2. \( \sum_{r=1}^{n} r^2 = \frac{n(n + 1)(2n + 1)}{6} \)

Substituting these formulas into the expression for \( S \):
\[ S = n \cdot \frac{n(n + 1)}{2} - \frac{n(n + 1)(2n + 1)}{6} + \frac{n(n + 1)}{2} \]

### Step 6: Combine the terms
Combining the terms gives:
\[ S = \frac{n(n + 1)}{2} \cdot n - \frac{n(n + 1)(2n + 1)}{6} + \frac{n(n + 1)}{2} \]
Factoring out \( \frac{n(n + 1)}{2} \):
\[ S = \frac{n(n + 1)}{2} \left( n + 1 - \frac{(2n + 1)}{3} \right) \]

### Step 7: Simplify the expression
To simplify the expression inside the parentheses:
\[ n + 1 - \frac{(2n + 1)}{3} = \frac{3(n + 1) - (2n + 1)}{3} = \frac{3n + 3 - 2n - 1}{3} = \frac{n + 2}{3} \]

### Step 8: Final expression for the sum
Substituting back, we have:
\[ S = \frac{n(n + 1)}{2} \cdot \frac{n + 2}{3} = \frac{n(n + 1)(n + 2)}{6} \]

Thus, the sum of the series \( n \cdot 1 + (n - 1) \cdot 2 + (n - 2) \cdot 3 + \ldots + 1 \) is:
\[ \boxed{\frac{n(n + 1)(n + 2)}{6}} \]