Final answer:
The sum of the first 30 terms of the series where each term is the product of squares of integers divided by the nth even number is 18630, using the sum of the first n squares formula.
Step-by-step explanation:
The given series has a pattern where each term is the product of squares of integers from 1 up to the integer n, divided by the nth even number. The nth term of the series can be expressed as (12 × 22 × ... × n2) / (2n). However, this problem simplifies when we notice a property: if you take (n − 1) from the last term and add it to the first term, you can pair the terms into sums of 2n, and it simplifies to 2n2. Therefore, the sum of 30 terms will be 2 × 12 + 2 × 22 + ... + 2 × 302.
Let's calculate a few initial terms to see the pattern:
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- 1st term: 2 × 12 = 2
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- 2nd term: 2 × 22 = 8
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- 3rd term: 2 × 32 = 18
Now we can continue this process for all 30 terms and sum them together.
But there's a shortcut, we can use the formula for the sum of the first n squares: S = n(n + 1)(2n + 1) / 6. Applying it to our series we get:
S = 2[(30)(31)(61) / 6] = 2[30 × 31 × 61 / 6]
After calculating the expression above, we find:
S = 2 × 30 × 31 × 61 / 6 = 18630
Therefore, the sum of the first 30 terms of the given series is 18630.