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What is the sum of the cubes of the first ten positive

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What is the sum of the cubes of the first ten positive  [#permalink]

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New post 12 Nov 2012, 20:00
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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. \(100^3\)
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Re: What is the sum of the cubes of the first ten positive integ  [#permalink]

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New post 12 Nov 2012, 20:53
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superpus07 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) \(100^3\)


Thanks!


The sum of the cubes of first n positive integers is given by \([n(n+1)/2]^2\)
If n = 10, the sum is 55^2.

Now, do you need to learn it up?
No. Even if you didn't know it, you should have tried to look at the pattern.
\(1^3 = 1 = 1^2\)
\(1^3 + 2^3 = 9 = 3^2\)
\(1^3 + 2^3 + 3^3 = 36 = 6^2\)
\(1^3 + 2^3 + 3^3 + 4^3 = 100 = 10^2\)
You see that when you find the sum of the numbers (not of the cubes) and square the sum, you get the sum of the cubes.
Sum of 1+2+3+4 = 10. Square it to get 100. Sum of \(1^3 + 2^3 + 3^3 + 4^3 = 100\)
This is pattern recognition.

You can also use another method - of averaging.
The numbers look like this: 1, 8, 27, 64, 125 ... etc
The numbers on the right 8^3, 9^3, 10^3 will be much larger than those on the left which are small. The average would lie not in the middle but on the right somewhere between 6^3 and 7^3. I would say around 300 so the sum will be around 300*10 = 3000. This leads us to 55^2.
The only hitch might be the 45^2 given which might make you uncomfortable in using this approximation. 45^2 = 2025

To ensure that avg is around 300 and not around 200, notice that from 10^3, you can make five 200s. From 9^3, you can make about four 200s (using some extra). So overall, you can make many more 200s than the required 10. Therefore, the average must be 300 and not 200.

Answer (C).
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Re: What is the sum of the cubes of the first ten positive  [#permalink]

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New post 13 Nov 2012, 17:45
Karishma, how did you get that formula [n(n+1)/2]^2?
Is that something we should be memorizing for the GMAT or is that a derived form of something else?
Can you provide guidance on other formulas that we should know? Thank you!!
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Re: What is the sum of the cubes of the first ten positive  [#permalink]

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New post 13 Nov 2012, 19:13
How does that pattern recognition help solve the problem? How does it translate to 5^3+6^3+7^3... to find 55^2 as the answer?
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Re: What is the sum of the cubes of the first ten positive  [#permalink]

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New post 13 Nov 2012, 19:34
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JJ2014 wrote:
Karishma, how did you get that formula [n(n+1)/2]^2?
Is that something we should be memorizing for the GMAT or is that a derived form of something else?
Can you provide guidance on other formulas that we should know? Thank you!!


There are quite a few related formulas:

Sum of first n consecutive positive integers: n(n+1)/2
Sum of first n consecutive positive odd integers: n^2
Sum of first n consecutive positive even integers: n(n+1)
Sum of squares of first n consecutive positive integers: n(n+1)(2n+1)/6
Sum of cubes of first n consecutive positive integers: [n(n+1)/2]^2

Do you need to memorize them for GMAT? No, except for the first one (since it is very useful). GMAT doesn't expect you to 'learn up' these formulas. The question will be such that you can use pattern recognition in such cases. But obviously it wouldn't hurt to know them.

I am no fan of memorizing tons of formulas. I concede that they are very useful in very few cases but I generally don't consider the probability of seeing those cases very high. But knowing these formulas can sometimes speed up your calculations. The question asks you something else - you recognize a pattern, you know a formula, you save time at the intermediate step and quickly arrive at the answer. I know these formulas because I have come across them many times so I ended up retaining them.

All in all, do I suggest you to learn them up? It's up to you. It's certainly not necessary. GMAT will not ask you a question where you must know these formulas. It's all about your commitment to the exam (if you want to leave no stone unturned, you might want to memorize them) and your aptitude (how easy it is for you to recognize patterns and derive a relation in exam conditions).
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Re: What is the sum of the cubes of the first ten positive  [#permalink]

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New post 13 Nov 2012, 19:38
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nlrosidi wrote:
How does that pattern recognition help solve the problem? How does it translate to 5^3+6^3+7^3... to find 55^2 as the answer?


Revisit the post above. You need to recognize that the first few sums obtained are all squares. 1, 9, 36, 100 ...
The square roots are 1, 3, 6, 10 ...
These are sum of first few numbers:
1 = 1
1 + 2 = 3
1 + 2 + 3 = 6
1 + 2 + 3 + 4 = 10

etc

Find the sum of first ten numbers: 1+ 2 + 3 + ...10 which is 10*11/2 = 55
Sum of cubes will be 55^2
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Re: What is the sum of the cubes of the first ten positive  [#permalink]

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New post 28 Jul 2014, 00:24
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VeritasPrepKarishma wrote:
superpus07 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) \(100^3\)


Thanks!


The sum of the cubes of first n positive integers is given by \([n(n+1)/2]^2\)
If n = 10, the sum is 55^2.

Now, do you need to learn it up?
No. Even if you didn't know it, you should have tried to look at the pattern.
\(1^3 = 1 = 1^2\)
\(1^3 + 2^3 = 9 = 3^2\)
\(1^3 + 2^3 + 3^3 = 36 = 6^2\)
\(1^3 + 2^3 + 3^3 + 4^3 = 100 = 10^2\)
You see that when you find the sum of the numbers (not of the cubes) and square the sum, you get the sum of the cubes.
Sum of 1+2+3+4 = 10. Square it to get 100. Sum of \(1^3 + 2^3 + 3^3 + 4^3 = 100\)
This is pattern recognition.

You can also use another method - of averaging.
The numbers look like this: 1, 8, 27, 64, 125 ... etc
The numbers on the right 8^3, 9^3, 10^3 will be much larger than those on the left which are small. The average would lie not in the middle but on the right somewhere between 6^3 and 7^3. I would say around 300 so the sum will be around 300*10 = 3000. This leads us to 55^2.
The only hitch might be the 45^2 given which might make you uncomfortable in using this approximation. 45^2 = 2025

To ensure that avg is around 300 and not around 200, notice that from 10^3, you can make five 200s. From 9^3, you can make about four 200s (using some extra). So overall, you can make many more 200s than the required 10. Therefore, the average must be 300 and not 200.

Answer (C).




I think this problem can be quickly solved by just looking at the answer choices.
(A) This can be eliminated since the answer obviously should be greater than 10^3 or 1000
(D) This can be eliminated since the answer should also definitely be less than 10*10^3 or less than 10000
(E) can be eliminated using similar logic for D above.

Now the contenders are 45^2 and 55^2.
Now 45^2 = 2025. and 55^2 = 3025. We know 10^3 = 1000 and 9^3 = 729. So we can easily see that the total of our required sum would exceed 2025. Hence answer = (C).

Hope this helps.
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Re: What is the sum of the cubes of the first ten positive  [#permalink]

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New post 17 Feb 2019, 00:39
isn't the question asking for the sum of the cubes of the first 10 intergers? which should be 1^3+2^3...+10^3, and the result should be 5005?
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Re: What is the sum of the cubes of the first ten positive  [#permalink]

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New post 08 Apr 2019, 09:37
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superpus07 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. \(100^3\)


Let's look for a pattern

1³ = 1 = 1²
1³ + 2³ = 9 = 3²
1³ + 2³ + 3³ = 36 = 6²
1³ + 2³ + 3³ + 4³ = 100 = 10²
1³ + 2³ + 3³ + 4³ + 5³ = 225 = 15²

See the pattern yet?
1 + 2 = 3
3 + 3 = 6
6 + 4 = 10
10 + 5 = 15

First we add 2, then we add 3, then 4, then 5, etc

So, continuing the pattern, we get:
1³ + 2³ + 3³ + . . . + 6³ = 21²
1³ + 2³ + 3³ + . . . + 7³ = 28²
1³ + 2³ + 3³ + . . . + 8³ = 36²
1³ + 2³ + 3³ + . . . + 9³ = 45²
1³ + 2³ + 3³ + . . . + 10³ = 55²

Answer: C

Cheers,
Brent
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Re: What is the sum of the cubes of the first ten positive   [#permalink] 08 Apr 2019, 09:37
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