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Answer:

answer is 106660789 OK zolved

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To find the sum of this series, you can recognize that it is an alternating series of squares. You can use the formula for the sum of an alternating series of squares:

S = 1² - 2² + 3² - 4² + ... + (-1)ⁿ × n²

In this series, n goes from 1 to 2009, and n² is squared. To simplify this, you can group the terms two by two:

S = (1² - 2²) + (3² - 4²) + ... + (2009² - 2010²)

Now, calculate each pair:

1² - 2² = -3

3² - 4² = -7

5² - 6² = -11

.....

2009² - 2010² = -4019

Now, you have a series of negative odd numbers. To find the sum of these numbers, you can use the formula for the sum of an arithmetic series:

[tex]{ \red{ \boxed{ \tt \: S = \frac{n}{2} \times [2a + (n - 1)d]}}}[/tex]

Where:

• n is the number of terms,
• a is the first term,
• d is the common difference.

In your case, n is 1005 (half of 2009), a is -3 (the first term), and d is -4 (the common difference between the terms). Plug these values into the formula:

[tex] \tt \: S = \dfrac{1005}{5} \times [2 \times (-3) + (1005 - 1) \times (-4)][/tex]

Now, calculate the sum:

S = 502.5 × (-6 - 4016)

S = 502.5 × (-4022)

[tex]{ \boxed{ \bold{S = -2,019,511}}}[/tex]

So, the sum of the given series is -2,019,511.

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[tex]{ \orange{ \sf \: hope \: it \: helps \: you}}[/tex]