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The sum of cubes of first five whole numbers

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The sum of cubes of first five whole numbers [#permalink] New post 20 Jul 2003, 00:15
The sum of cubes of first five whole numbers 1^3+2^3+...+5^3=225. Find the sum of cubes of the next five integers 6^3+7^3+...+10^3.
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Re: something like Akamaibrah's [#permalink] New post 20 Jul 2003, 02:09
stolyar wrote:
The sum of cubes of first five whole numbers 1^3+2^3+...+5^3=225. Find the sum of cubes of the next five integers 6^3+7^3+...+10^3.


Hmmm.

Challenge: Solve this without looking up or using the sum of cubes formula (using formula not fair because this is NOT knowledge ETS expects you to have). You CAN solve this without using or deriving the formula, or without resorting to actually computing the numbers.

Good luck!
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 [#permalink] New post 20 Jul 2003, 08:23
Okies,

My way is to split up each into 2 parts or 3 parts.

a^3 + b ^3 = (a+b)(a^2 + b ^2 - ab)

and a^3 + b^3 + c^3 = (a+b+c)(a^2+b^2+c^2 - ab - ac - bc) + 3abc.

Goingby that i split up, 6, 7 & 8 as a part and 9 and 10 as other part.

Got the answer 2800.

If this formula is also not allowed then what are we expected to know...

Anyways i shall try to think of something more better.... :roll:
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 [#permalink] New post 20 Jul 2003, 09:08
A sum of cubes of 1, 2, 3, .... , n is (n^2)*(n+1)^2/4

Memorize it, it is simple and helpful.
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 [#permalink] New post 20 Jul 2003, 10:52
evensflow wrote:
Okies,

My way is to split up each into 2 parts or 3 parts.

a^3 + b ^3 = (a+b)(a^2 + b ^2 - ab)

and a^3 + b^3 + c^3 = (a+b+c)(a^2+b^2+c^2 - ab - ac - bc) + 3abc.

Goingby that i split up, 6, 7 & 8 as a part and 9 and 10 as other part.

Got the answer 2800.

If this formula is also not allowed then what are we expected to know...

Anyways i shall try to think of something more better.... :roll:


ONe good general method (the one I used) is to express each term as (5 + 1)^3, (5 + 2)^3, then expand each term. To save time, I will give you (5 + x)^3 = (125 + 75 x + 15 x^2 + x^3). Going term by term, you now have a bunch of sums that are easy to sum up as "x" goes from 1 to 5 -- try it. (BTW, sum of 1st 5 squares is 55).

5* 125
5 * 75 * (1 + 5)/2
15 (1 + 4 + 9 + 16 + 25)
225
= 2800

Stolyar wrote:
A sum of cubes of 1, 2, 3, .... , n is (n^2)*(n+1)^2/4

Memorize it, it is simple and helpful.


If course if you know the formula, it is easy. But is it smart?

I am 100% against advice to learn yet another formula, especially one that we are not expected to know. I NEVER teach the use of "special-case" formulas, not even the one for Combinations. After all, the combination formula is simply the permutation formula with one "adjustment" made WHEN APPROPRIATE. I go crazy when I see people bandy terms like Binomial Expansion and Hypergeometric distribution formulas, as though their knowledge of these formulas make them "cool." Yeah, the formula works, but it doesn't promote the understanding of what is really going on.

The GMAT is not a test about knowing lots of specialized math. It is a test of one's ability to solve a huge variety of problems equipped with only a few basic principles, a systematic approach to problem solving, and logic, imagination, creativity, and out-of-the-box thinking. Encouraging the memorization of yet another esoteric formula as a way to gain an edge is, IMO, counterproductive and BAD advice. Exactly which of the infinite number of potentially helpful specialized formulas should we learn? And once we create a dependency of the application of a special formula, what if one forgots the formula, or worse, the situation is such the the formula that one had taken great pains to memorize doesn quite apply?

The difference between the 690 guy and the 790 guy, IMO, is not the difference in their arsenals of memorized formulas, but the ability of the 790 guy to take the few principles that ETS requires us to know, and creatively apply them to anything ETS can dish out. THAT is how the very best score high on the Q side.
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 [#permalink] New post 21 Jul 2003, 15:47
i pefectly agree with you Akamai that mugging just formulaes wont help. I belive that there are n ways to solve a math problem. It just depends upon the comfort of a person, who has to decide which way is best for him.

I didnt know the formula for the sum of cubes. So, i didnt even have to see your post first, which said dont use formulaes.

Later, i tried to figure out the formula of the sum of the cubes by myself and figured out that it almost like sum to first n natural numbers [i.e. n(n+1)/2 ]. (Sum of cubes therefore is n^2(n+1)^2/4 , effectively squaring the sum of first n natural numbers).

Another thing Stoluyar posted it for me :) .

So yes i would agree that formulaes wont help on the exam day since there is a lot of chance that one might forget it admist all the tension. But, sometimes it can save a valuable second or two without having to stress too much on alternatives. But this is only for those who wish to keep 690 :wink:

Did i say a lot... i actually messed quite a bit on my test... :(
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 [#permalink] New post 21 Jul 2003, 16:13
evensflow wrote:
i pefectly agree with you Akamai that mugging just formulaes wont help. I belive that there are n ways to solve a math problem. It just depends upon the comfort of a person, who has to decide which way is best for him.

I didnt know the formula for the sum of cubes. So, i didnt even have to see your post first, which said dont use formulaes.

Later, i tried to figure out the formula of the sum of the cubes by myself and figured out that it almost like sum to first n natural numbers [i.e. n(n+1)/2 ]. (Sum of cubes therefore is n^2(n+1)^2/4 , effectively squaring the sum of first n natural numbers).

Another thing Stoluyar posted it for me :) .

So yes i would agree that formulaes wont help on the exam day since there is a lot of chance that one might forget it admist all the tension. But, sometimes it can save a valuable second or two without having to stress too much on alternatives. But this is only for those who wish to keep 690 :wink:

Did i say a lot... i actually messed quite a bit on my test... :(


I is so tempting to say that learning a particular formula is good, especially if it help you solve a particular problem. IMO, a strategy of learning formulas to help your GMAT is counterproductive, and offering a specialized formula as a solution for a GMAT problem is just bad practice.

Do you play golf? If you do, most players expect to have a putt of 2 feet or less "given" to them. If you make them putt it unexpectedly, they get all bent out of shape and these people miss the putt a disproportionate number of times. I NEVER accept gimmes, because by never accepting them, that gives me the mental attitude that I always have to sink the putt, therefore, I am never surprised that I have to actually putt the damn thing and I make them virtually every time.

Accepting the gimme is like accepting a formula. It helps you for this particular problem, but it doesn't help your general putting or problem solving skill in general and gives you a poor mind set.

I KNOW that ETS composes every problem so that it is solvable without using a formula, so I look at EVERY problem with an open mind, knowing that there is a nice insight that will lead me to the solution without esoteric mathematical knowledge. Hence, I apply the right attitude towards solving these problems BY HABIT and am very successful as a result.

I have found that my students with that same type of "non formula dependent" mindset solve complex GMAT problem far more often than those with more mathematical training who have grown accustomed to using the perfect formula.

In addition, learning the "non-formula" solutions to problems enhances your toolkit of analytical possibilities for other problems of this sort. IMO, this approach improves your general problems solving abilities much more that learning yet another formula does.

Given the choice, I prefer to teach my students how to THINK.
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 [#permalink] New post 21 Jul 2003, 16:19
Although, i dont know much about golf :wink: , but i got your point.

And i definitely agree with you :)

No two ways ......
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 [#permalink] New post 11 Oct 2005, 06:46
Hi....altho' this is a very old post, bringing it up again.....

so just was not clear on the formula for (5+x)^3 written by AkamaiBrah....how did 75 and 15 come about....

(5 + x)^3 = (125 + 75 x + 15 x^2 + x^3).

Anyone?
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  [#permalink] 11 Oct 2005, 06:46
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