The sum of the series `1^(2)+1+2^(2)+2+3^(2)+3+ . . . . .. +n^(2)+n`, is

Options

A. `(n(n+1))/(2)`
B. `{(n(+1))/(2)}^(2)`
C. `(n(n+1)(n+2))/(3)`
D. `(n(n+1)(n+2)(n+3))/(4)`

Answer

To find the sum of the series \( S = 1^2 + 1 + 2^2 + 2 + 3^2 + 3 + \ldots + n^2 + n \), we can rearrange the series into two separate sums: the sum of the squares and the sum of the integers.

### Step 1: Rearranging the Series
We can express the series as:
\[ S = (1^2 + 2^2 + 3^2 + \ldots + n^2) + (1 + 2 + 3 + \ldots + n) \]

### Step 2: Using the Formula for the Sum of Integers
The sum of the first \( n \) integers is given by the formula:
\[ \text{Sum of integers} = \frac{n(n + 1)}{2} \]

### Step 3: Using the Formula for the Sum of Squares
The sum of the squares of the first \( n \) integers is given by the formula:
\[ \text{Sum of squares} = \frac{n(n + 1)(2n + 1)}{6} \]

### Step 4: Combining the Sums
Now we can substitute the formulas into our expression for \( S \):
\[ S = \frac{n(n + 1)(2n + 1)}{6} + \frac{n(n + 1)}{2} \]

### Step 5: Finding a Common Denominator
To combine these two fractions, we need a common denominator. The common denominator is 6:
\[ S = \frac{n(n + 1)(2n + 1)}{6} + \frac{3n(n + 1)}{6} \]

### Step 6: Combining the Numerators
Now we can combine the numerators:
\[ S = \frac{n(n + 1)(2n + 1 + 3)}{6} \]
\[ S = \frac{n(n + 1)(2n + 4)}{6} \]

### Step 7: Simplifying the Expression
We can factor out a 2 from \( (2n + 4) \):
\[ S = \frac{n(n + 1) \cdot 2(n + 2)}{6} \]
\[ S = \frac{n(n + 1)(n + 2)}{3} \]

### Final Result
Thus, the sum of the series \( S \) is:
\[ S = \frac{n(n + 1)(n + 2)}{3} \]