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The sum of the squares of the first 15 positive integers (12 + 22 + 32 + . . . + 152) is equal to 1240. What is the sum of the squares of the second 15 positive integers (162 + 172 + 182 + . . . + 302) ?
- a)2480
- b)3490
- c)6785
- d)8215
- e)9255
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
The sum of the squares of the first 15 positive integers (12 + 22 + 32...
The key to solving this problem is to represent the sum of the squares of the second 15 integers as follows: (15 + 1)2 + (15 + 2)2 + (15 + 3)2 + . . . + (15 + 15)2.
Recall the popular quadratic form, (a + b)2 = a2 + 2ab + b2. Construct a table that uses this expansion to calculate each component of each term in the series as follows:
In order to calculate the desired sum, we can find the sum of each of the last 3 columns and then add these three subtotals together. Note that since each column follows a simple pattern, we do not have to fill in the whole table, but instead only need to calculate a few terms in order to determine the sums.
The column labeled a2 simply repeats 225 fifteen times; therefore, its sum is 15(225) = 3375.
The column labeled 2ab is an equally spaced series of positive numbers. Recall that the average of such a series is equal to the average of its highest and lowest values; thus, the average term in this series is (30 + 450) / 2 = 240. Since the sum of n numbers in an equally spaced series is simply n times the average of the series, the sum of this series is 15(240) = 3600.
The last column labeled b2 is the sum of the squares of the first 15 integers. This was given to us in the problem as 1240.
Finally, we sum the 3 column totals together to find the sum of the squares of the second 15 integers: 3375 + 3600 + 1240 = 8215. The correct answer choice is (D).
Most Upvoted Answer
The sum of the squares of the first 15 positive integers (12 + 22 + 32...
To find the sum of the squares of the second 15 positive integers, we can use the formula for the sum of squares of consecutive positive integers:
Sum of squares = (n * (n + 1) * (2n + 1)) / 6
where n is the last positive integer in the series. In this case, n = 15.
Let's calculate the sum of squares for the second 15 positive integers step by step:
1. Find the sum of squares for the first 15 positive integers:
Sum of squares = (15 * (15 + 1) * (2*15 + 1)) / 6
= (15 * 16 * 31) / 6
= 1240
2. Find the sum of squares for the second 15 positive integers:
Since the series starts from 16, we need to subtract the sum of squares of the first 15 positive integers from the sum of squares of the first 30 positive integers.
Sum of squares for the second 15 positive integers = (Sum of squares for the first 30 positive integers) - (Sum of squares for the first 15 positive integers)
= (30 * (30 + 1) * (2*30 + 1)) / 6 - (15 * (15 + 1) * (2*15 + 1)) / 6
= (30 * 31 * 61) / 6 - (15 * 16 * 31) / 6
= (30 * 31 * 61 - 15 * 16 * 31) / 6
= (31 * 31 * 30 - 31 * 16 * 15) / 6
= (31 * (31 * 30 - 16 * 15)) / 6
= (31 * (930 - 240)) / 6
= (31 * 690) / 6
= 10710 / 6
= 1785
Therefore, the sum of the squares of the second 15 positive integers is 1785.
Answer: Option D) 8215
Sum of squares = (n * (n + 1) * (2n + 1)) / 6
where n is the last positive integer in the series. In this case, n = 15.
Let's calculate the sum of squares for the second 15 positive integers step by step:
1. Find the sum of squares for the first 15 positive integers:
Sum of squares = (15 * (15 + 1) * (2*15 + 1)) / 6
= (15 * 16 * 31) / 6
= 1240
2. Find the sum of squares for the second 15 positive integers:
Since the series starts from 16, we need to subtract the sum of squares of the first 15 positive integers from the sum of squares of the first 30 positive integers.
Sum of squares for the second 15 positive integers = (Sum of squares for the first 30 positive integers) - (Sum of squares for the first 15 positive integers)
= (30 * (30 + 1) * (2*30 + 1)) / 6 - (15 * (15 + 1) * (2*15 + 1)) / 6
= (30 * 31 * 61) / 6 - (15 * 16 * 31) / 6
= (30 * 31 * 61 - 15 * 16 * 31) / 6
= (31 * 31 * 30 - 31 * 16 * 15) / 6
= (31 * (31 * 30 - 16 * 15)) / 6
= (31 * (930 - 240)) / 6
= (31 * 690) / 6
= 10710 / 6
= 1785
Therefore, the sum of the squares of the second 15 positive integers is 1785.
Answer: Option D) 8215
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The sum of the squares of the first 15 positive integers (12 + 22 + 32 + . . . + 152) is equal to 1240. What is the sum of the squares of the second 15 positive integers (162 + 172 + 182 + . . . + 302) ?a)2480b)3490c)6785d)8215e)9255Correct answer is option 'D'. Can you explain this answer? for GMAT 2025 is part of GMAT preparation. The Question and answers have been prepared according to the GMAT exam syllabus. Information about The sum of the squares of the first 15 positive integers (12 + 22 + 32 + . . . + 152) is equal to 1240. What is the sum of the squares of the second 15 positive integers (162 + 172 + 182 + . . . + 302) ?a)2480b)3490c)6785d)8215e)9255Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for GMAT 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The sum of the squares of the first 15 positive integers (12 + 22 + 32 + . . . + 152) is equal to 1240. What is the sum of the squares of the second 15 positive integers (162 + 172 + 182 + . . . + 302) ?a)2480b)3490c)6785d)8215e)9255Correct answer is option 'D'. Can you explain this answer?.
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