`(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 problem `(1^2 - 2^2 + 3^2) + (4^2 - 5^2 + 6^2) + (7^2 - 8^2 + 9^2) + ...` up to 30 terms, we can break it down step by step.

### Step 1: Identify the pattern in the series
The series can be grouped as follows:
- The first group is \(1^2 - 2^2 + 3^2\)
- The second group is \(4^2 - 5^2 + 6^2\)
- The third group is \(7^2 - 8^2 + 9^2\)
- This pattern continues.

Each group consists of three terms, where the first term is positive, the second term is negative, and the third term is positive.

### Step 2: Determine the number of groups
Since we have 30 terms, and each group contains 3 terms, we can find the number of groups:
\[ \text{Number of groups} = \frac{30}{3} = 10 \]

### Step 3: Write the general form of each group
The general form of each group can be expressed as:
\[ (3n - 2)^2 - (3n - 1)^2 + (3n)^2 \]
where \(n\) is the group number (from 1 to 10).

### Step 4: Expand each group
Using the difference of squares formula \(a^2 - b^2 = (a-b)(a+b)\), we can expand each group:
1. For \(n = 1\):
\[ (1^2 - 2^2 + 3^2) = (1 - 2)(1 + 2) + 3^2 = (-1)(3) + 9 = -3 + 9 = 6 \]
2. For \(n = 2\):
\[ (4^2 - 5^2 + 6^2) = (4 - 5)(4 + 5) + 6^2 = (-1)(9) + 36 = -9 + 36 = 27 \]
3. For \(n = 3\):
\[ (7^2 - 8^2 + 9^2) = (7 - 8)(7 + 8) + 9^2 = (-1)(15) + 81 = -15 + 81 = 66 \]
4. Continuing this pattern, we can see that:
\[ \text{For } n = 10: (30^2 - 31^2 + 32^2) = (30 - 31)(30 + 31) + 32^2 = (-1)(61) + 1024 = -61 + 1024 = 963 \]

### Step 5: Sum all groups
Now we need to sum all the results from each group:
\[ S = 6 + 27 + 66 + ... + 963 \]
We can observe that the results form an arithmetic series.

### Step 6: Calculate the total
To find the sum of the arithmetic series, we can use the formula for the sum of an arithmetic series:
\[ S_n = \frac{n}{2} \times (a + l) \]
where:
- \(n\) is the number of terms (10),
- \(a\) is the first term (6),
- \(l\) is the last term (963).

Calculating:
\[ S_{10} = \frac{10}{2} \times (6 + 963) = 5 \times 969 = 4845 \]

### Final Answer
Thus, the sum of the series up to 30 terms is:
\[ \boxed{4845} \]