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Re: What is the sum of the cubes of the first ten positive integers? [#permalink]
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Sum of cubes of first n natural numbers = \((n(n+1)/2)^2\)

Sum of cubes of first 10 positive integers =\((10(10+1)/2)^2\) = \((55)^2\)

Ans : C
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Re: What is the sum of the cubes of the first ten positive integers? [#permalink]
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Bunuel wrote:
What is the sum of the cubes of the first ten positive integers?

(A) 10^3
(B) 45^2
(C) 55^2
(D) 100^2
(E) 100^3

Kudos for a correct solution.


MANHATTAN GMAT OFFICIAL SOLUTION:

Set up a table to list the first few cubes and track the cumulative sum:
Attachment:
2015-06-15_1416.png
2015-06-15_1416.png [ 18.72 KiB | Viewed 14042 times ]

By this point, you should start to recognize that the cumulative sums are all perfect squares. Extend the table to the right to list the square roots.
Attachment:
2015-06-15_1416_001.png
2015-06-15_1416_001.png [ 57.05 KiB | Viewed 13953 times ]

Not only do we see a pattern emerging for the cumulative sum (always a perfect square), but also for the square root of the cumulative sum, which is always the sum of consecutive integers.

Thus, the sum of the first 10 cubes will be the square of the sum of the first 10 positive integers:
1^3 + 2^3 + 3^3 + 4^3 + … + 10^3 = (1 + 2 + 3 + 4 + … + 10)^2

Now, the sum of the first 10 positive integers can be computed by matching pairs:
1 + 2 + … + 9 + 10 = 11 + 11 + 11 + 11 + 11 = 5(11) = 55.

So, the sum of the first 10 cubes is 55^2.

The correct answer is C.
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Re: What is the sum of the cubes of the first ten positive integers? [#permalink]
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