What is the sum of n terms of the series `-1 + 1^2 - 2 + 2^2 - 3 + 3^2` … ?

Options

A. `(n(n^2 +1))/(3)`
B. `n^2 +n `
C. `(n(n^2 -1))/(3)`
D. none of these

Answer

To find the sum of the first n terms of the series `-1 + 1^2 - 2 + 2^2 - 3 + 3^2 ...`, we can analyze the series step by step.

### Step 1: Identify the Pattern
The series alternates between negative integers and the squares of positive integers:
- The first term is `-1`
- The second term is `1^2 = 1`
- The third term is `-2`
- The fourth term is `2^2 = 4`
- The fifth term is `-3`
- The sixth term is `3^2 = 9`

This pattern continues, where the odd-indexed terms are negative integers and the even-indexed terms are squares of integers.

### Step 2: Group the Terms
We can group the terms in pairs:
- First pair: `-1 + 1^2 = -1 + 1 = 0`
- Second pair: `-2 + 2^2 = -2 + 4 = 2`
- Third pair: `-3 + 3^2 = -3 + 9 = 6`

Notice that each pair contributes a certain value to the sum.

### Step 3: Generalize the Sum of Pairs
For the k-th pair, the sum can be expressed as:
- Sum of k-th pair: `-k + k^2 = k^2 - k`

### Step 4: Calculate the Sum of n Terms
If n is even, we can express n as `n = 2m` (where m is the number of pairs):
- The sum of the first n terms will be the sum of the first m pairs:

\[ S_n = \sum_{k=1}^{m} (k^2 - k) \]

This can be simplified:
\[ S_n = \sum_{k=1}^{m} k^2 - \sum_{k=1}^{m} k \]

Using the formulas for the sums:
- Sum of the first m squares: \(\frac{m(m + 1)(2m + 1)}{6}\)
- Sum of the first m natural numbers: \(\frac{m(m + 1)}{2}\)

Thus, we can write:
\[ S_n = \frac{m(m + 1)(2m + 1)}{6} - \frac{m(m + 1)}{2} \]

### Step 5: Simplify the Expression
Combining the two terms:
\[ S_n = \frac{m(m + 1)(2m + 1)}{6} - \frac{3m(m + 1)}{6} = \frac{m(m + 1)(2m + 1 - 3)}{6} \]
\[ = \frac{m(m + 1)(2m - 2)}{6} = \frac{m(m + 1)(2(m - 1))}{6} \]
\[ = \frac{m(m + 1)(m - 1)}{3} \]

### Step 6: Final Result
Thus, the sum of the first n terms of the series is:
- If n is even, \(S_n = \frac{m(m + 1)(m - 1)}{3}\) where \(m = \frac{n}{2}\).