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16 Sep 2014, 01:52
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C
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Difficulty:
45%
(medium)
Question Stats:
72% (02:48) correct
28%
(02:53)
wrong
based on 71
sessions
History
Date
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The sum of the first \(n\) positive perfect squares, where \(n\) is a positive integer, is given by the formula \(\frac{n^3}{3} + cn^2 + \frac{n}{6}\), where \(c\) is a constant. What is the sum of the first 15 positive perfect squares?
A. 1010
B. 1164
C. 1240
D. 1316
E. 1476
A. 1010
B. 1164
C. 1240
D. 1316
E. 1476
16 Sep 2014, 01:53
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Official Solution:
The sum of the first \(n\) positive perfect squares, where \(n\) is a positive integer, is given by the formula \(\frac{n^3}{3} + cn^2 + \frac{n}{6}\), where \(c\) is a constant. What is the sum of the first 15 positive perfect squares?
A. 1010
B. 1164
C. 1240
D. 1316
E. 1476
The brute-force way to solve this problem is literally to add up the first 15 positive perfect squares, from 1 to 225, inclusive. This is not necessarily completely out of bounds, given that we only have to sum up 15 numbers, all of which we should know already, and several of which are small. However, we should look for a shortcut using the formula.
Unfortunately, there is an unknown constant in the formula, but by using a small test number, we can solve for this constant. You can certainly pick \(n = 1\), since it is a positive integer:
\(1^2 = \frac{1^3}{3} + c1^2 + \frac{1}{6}\)
\(1 = \frac{1}{3} + c + \frac{1}{6}\)
\(\frac{1}{2} = c\)
If you feel uncomfortable picking \(n = 1\), you can pick \(n = 2\) and come to the same result almost as quickly.
Now, we plug \(n = 15\) into the formula and solve:
\(1^2 + 2^2 + ... + 15^2 = \frac{15^3}{3} + \frac{15^2}{2} + \frac{15}{6} =\)
\(\frac{15 \times 15 \times 15}{3} + \frac{15 \times 15}{2} + \frac{15}{6} =\)
\(15 \times 15 \times 5 + \frac{15 \times 15}{2} + \frac{5}{2} =\)
\(225 \times 5 + \frac{225}{2} + \frac{5}{2} =\)
\(1125 + \frac{230}{2} =\)
\(1125 + 115 =\)
\(= 1240\)
Answer: C
The sum of the first \(n\) positive perfect squares, where \(n\) is a positive integer, is given by the formula \(\frac{n^3}{3} + cn^2 + \frac{n}{6}\), where \(c\) is a constant. What is the sum of the first 15 positive perfect squares?
A. 1010
B. 1164
C. 1240
D. 1316
E. 1476
The brute-force way to solve this problem is literally to add up the first 15 positive perfect squares, from 1 to 225, inclusive. This is not necessarily completely out of bounds, given that we only have to sum up 15 numbers, all of which we should know already, and several of which are small. However, we should look for a shortcut using the formula.
Unfortunately, there is an unknown constant in the formula, but by using a small test number, we can solve for this constant. You can certainly pick \(n = 1\), since it is a positive integer:
\(1^2 = \frac{1^3}{3} + c1^2 + \frac{1}{6}\)
\(1 = \frac{1}{3} + c + \frac{1}{6}\)
\(\frac{1}{2} = c\)
If you feel uncomfortable picking \(n = 1\), you can pick \(n = 2\) and come to the same result almost as quickly.
Now, we plug \(n = 15\) into the formula and solve:
\(1^2 + 2^2 + ... + 15^2 = \frac{15^3}{3} + \frac{15^2}{2} + \frac{15}{6} =\)
\(\frac{15 \times 15 \times 15}{3} + \frac{15 \times 15}{2} + \frac{15}{6} =\)
\(15 \times 15 \times 5 + \frac{15 \times 15}{2} + \frac{5}{2} =\)
\(225 \times 5 + \frac{225}{2} + \frac{5}{2} =\)
\(1125 + \frac{230}{2} =\)
\(1125 + 115 =\)
\(= 1240\)
Answer: C
27 Jan 2015, 06:32
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Hey,
One question: why did you square 1, in your test formula to find c? Why isn't it just 1? Shouldn't it be n=the given formula?
One question: why did you square 1, in your test formula to find c? Why isn't it just 1? Shouldn't it be n=the given formula?
