Find the sum to `n` terms of the series `1^2+2^2+3^2-4^2+5^2-6^2+. . . .`

Explanation

To find the sum to `n` terms of the series \(1^2 + 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + \ldots\), we first need to analyze the pattern in the series.

### Step 1: Identify the Pattern
The series alternates between positive and negative squares:
- The odd-indexed terms (1, 3, 5, ...) are positive.
- The even-indexed terms (2, 4, 6, ...) are negative.

### Step 2: Separate Cases for Even and Odd `n`
We will consider two cases based on whether `n` is even or odd.

#### Case 1: `n` is Even
Let \( n = 2k \) for some integer \( k \). The series can be grouped as follows:
\[ (1^2 - 2^2) + (3^2 - 4^2) + (5^2 - 6^2) + \ldots + ((2k-1)^2 - (2k)^2) \]
Using the difference of squares, \( a^2 - b^2 = (a - b)(a + b) \):
\[ = (1 - 2)(1 + 2) + (3 - 4)(3 + 4) + (5 - 6)(5 + 6) + \ldots + ((2k-1) - (2k))((2k-1) + (2k)) \]
This simplifies to:
\[ = -1 \cdot 3 + -1 \cdot 7 + -1 \cdot 11 + \ldots + -1 \cdot (4k - 1) \]
The sum of the first \( k \) odd numbers is \( k^2 \), so:
\[ = -1 \cdot (3 + 7 + 11 + \ldots + (4k - 1)) \]
The series \( 3, 7, 11, \ldots, (4k - 1) \) is an arithmetic series with first term 3, common difference 4, and \( k \) terms. The sum of an arithmetic series is given by:
\[ \text{Sum} = \frac{k}{2} \cdot (\text{first term} + \text{last term}) = \frac{k}{2} \cdot (3 + (4k - 1)) = \frac{k}{2} \cdot (4k + 2) = k(2k + 1) \]
Thus, the sum for even \( n \) is:
\[ S_n = -k(2k + 1) = -\frac{n(n + 1)}{4} \]

#### Case 2: `n` is Odd
Let \( n = 2k + 1 \). The series becomes:
\[ (1^2 - 2^2) + (3^2 - 4^2) + (5^2 - 6^2) + \ldots + ((2k-1)^2 - (2k)^2) + (2k + 1)^2 \]
The first part is the same as in the even case:
\[ = -k(2k + 1) + (2k + 1)^2 \]
Calculating \( (2k + 1)^2 \):
\[ = 4k^2 + 4k + 1 \]
Thus, the total sum for odd \( n \) is:
\[ S_n = -k(2k + 1) + (4k^2 + 4k + 1) = 4k^2 + 4k + 1 - (2k^2 + k) = 2k^2 + 3k + 1 \]
Substituting \( k = \frac{n - 1}{2} \):
\[ S_n = 2\left(\frac{n - 1}{2}\right)^2 + 3\left(\frac{n - 1}{2}\right) + 1 = \frac{(n - 1)^2}{2} + \frac{3(n - 1)}{2} + 1 \]
Simplifying:
\[ = \frac{n^2 - 2n + 1 + 3n - 3 + 2}{2} = \frac{n^2 + n}{2} \]

### Final Result
Thus, the sum of the series to `n` terms is:
\[ S_n = \begin{cases} -\frac{n(n + 1)}{4} & \text{if } n \text{ is even} \\ \frac{n(n + 1)}{2} & \text{if } n \text{ is odd} \end{cases} \]