The sum of the following series
`(1^2 +1) + (2^2 + 2) + (3^2 + 3) + (4^2 + 4) + …..+(n^2 + n)` is :

Options

A. `a)n^3`
B. b)`((n^2 + n)5)/(3)`
C. c)`(n(n + 1)(n+2))/(3)`
D. d)can't be determined

Answer

To find the sum of the series \( (1^2 + 1) + (2^2 + 2) + (3^2 + 3) + (4^2 + 4) + \ldots + (n^2 + n) \), we can break it down into two separate summations:

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

### Step 1: Write the series in summation form
We can express the series as:
\[ \sum_{k=1}^{n} (k^2 + k) = \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k \]

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

### Step 3: Substitute the formulas into the expression
Substituting these formulas into our expression, we have:
\[ \sum_{k=1}^{n} (k^2 + k) = \frac{n(n + 1)(2n + 1)}{6} + \frac{n(n + 1)}{2} \]

### Step 4: Find a common denominator and simplify
To combine these two fractions, we need a common denominator, which is 6:
\[ \frac{n(n + 1)(2n + 1)}{6} + \frac{3n(n + 1)}{6} = \frac{n(n + 1)(2n + 1 + 3)}{6} \]
This simplifies to:
\[ \frac{n(n + 1)(2n + 4)}{6} \]

### Step 5: Factor out the common terms
We can factor out a 2 from \( (2n + 4) \):
\[ \frac{n(n + 1) \cdot 2(n + 2)}{6} = \frac{n(n + 1)(n + 2)}{3} \]

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