`(1^(2)-2^(2)+3^(2)) + (4^(2)-5^(2)+6^(2))+(7^(2)-8^(2)+9^(2))….30` terms

Options

A. 82395
B. 80396
C. 82389
D. 82399

Explanation

To solve the given expression \((1^{2}-2^{2}+3^{2}) + (4^{2}-5^{2}+6^{2})+(7^{2}-8^{2}+9^{2})+\ldots\) up to 30 terms, we can break it down step by step.

### Step 1: Identify the Pattern
The series can be grouped as follows:
- The first group: \(1^2 - 2^2 + 3^2\)
- The second group: \(4^2 - 5^2 + 6^2\)
- The third group: \(7^2 - 8^2 + 9^2\)
- And so on...

Each group consists of three terms. The first term is a perfect square, the second term is subtracted, and the third term is added.

### Step 2: Calculate Each Group
Using the difference of squares formula, \(a^2 - b^2 = (a-b)(a+b)\), we can simplify each group.

For the first group:
\[ 1^2 - 2^2 + 3^2 = 1 - 4 + 9 = 6 \]

For the second group:
\[ 4^2 - 5^2 + 6^2 = 16 - 25 + 36 = 27 \]

For the third group:
\[ 7^2 - 8^2 + 9^2 = 49 - 64 + 81 = 66 \]

### Step 3: Generalize the Group Calculation
The \(n\)-th group can be represented as:
\[ (3n-2)^2 - (3n-1)^2 + (3n)^2 \]
Calculating this:
\[ = (3n-2)^2 - (3n-1)^2 + (3n)^2 = [(3n-2) - (3n-1)][(3n-2) + (3n-1)] + (3n)^2 = (-1)(6n - 3) + 9n^2 = -6n + 3 + 9n^2 = 9n^2 - 6n + 3 \]

### Step 4: Sum the Groups
Now, we need to sum this expression for \(n\) from 1 to 10 (since there are 30 terms and each group has 3 terms):
\[ \sum_{n=1}^{10} (9n^2 - 6n + 3) \]

### Step 5: Calculate Each Part of the Sum
1. **Sum of \(n^2\)**:
\[ \sum_{n=1}^{10} n^2 = \frac{10(10 + 1)(2 \cdot 10 + 1)}{6} = \frac{10 \cdot 11 \cdot 21}{6} = 385 \]

2. **Sum of \(n\)**:
\[ \sum_{n=1}^{10} n = \frac{10(10 + 1)}{2} = 55 \]

3. **Constant Sum**:
\[ \sum_{n=1}^{10} 3 = 3 \cdot 10 = 30 \]

### Step 6: Combine the Results
Now substituting back:
\[ = 9 \cdot 385 - 6 \cdot 55 + 30 = 3465 - 330 + 30 = 3465 - 300 = 3165 \]

### Final Answer
The sum of the series is \(3165\).