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

Options

A. `(n(n +1)(n+2))/(3)`
B. `(n(n+1) (3n +1))/(2)`
C. `(n(2n+1)(3n +1))/(6)`
D. `[(n( +1))/(3)]^(2)`

Answer

To find the sum of the series \(1^2 + 1 + 2^2 + 2 + 3^2 + 3 + \ldots + n^2 + n\), we can break this series into two parts:

1. The sum of the squares: \(1^2 + 2^2 + 3^2 + \ldots + n^2\)
2. The sum of the integers: \(1 + 2 + 3 + \ldots + n\)

### Step 1: Sum of the squares

The formula for the sum of the squares of the first \(n\) natural numbers is given by:

\[ \text{Sum of squares} = \frac{n(n + 1)(2n + 1)}{6} \]

### Step 2: Sum of the integers

The formula for the sum of the first \(n\) natural numbers is:

\[ \text{Sum of integers} = \frac{n(n + 1)}{2} \]

### Step 3: Combine the two sums

Now, we can combine these two results to get the total sum of the series:

\[ \text{Total Sum} = \left(\frac{n(n + 1)(2n + 1)}{6}\right) + \left(\frac{n(n + 1)}{2}\right) \]

### Step 4: Find a common denominator

To combine these two fractions, we need a common denominator. The least common multiple of 6 and 2 is 6. We can rewrite the second term:

\[ \frac{n(n + 1)}{2} = \frac{3n(n + 1)}{6} \]

### Step 5: Combine the fractions

Now we can add the two fractions:

\[ \text{Total Sum} = \frac{n(n + 1)(2n + 1)}{6} + \frac{3n(n + 1)}{6} \]

Combining these gives:

\[ \text{Total Sum} = \frac{n(n + 1)(2n + 1 + 3)}{6} = \frac{n(n + 1)(2n + 4)}{6} \]

### Step 6: Simplify the expression

We can factor out a 2 from the expression \(2n + 4\):

\[ \text{Total Sum} = \frac{n(n + 1)(2(n + 2))}{6} \]

This simplifies to:

\[ \text{Total Sum} = \frac{n(n + 1)(n + 2)}{3} \]

### Final Answer

Thus, the sum of the series \(1^2 + 1 + 2^2 + 2 + 3^2 + 3 + \ldots + n^2 + n\) is:

\[ \frac{n(n + 1)(n + 2)}{3} \]