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What is the Sum of 1^3+2^3+3^3+4^3+5^3+6^3+7^3

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alphabeta1234
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What is the Sum of 1^3+2^3+3^3+4^3+5^3+6^3+7^3 [#permalink] What is the Sum of 1^3+2^3+3^3+4^3+5^3+6^3+7^3   27 Jul 2013, 14:14
What is the sum of series S? Is there a shortcut/formula? I understand you can evaluate and sum each expression but I am trying to find a better solution.

S=1^3+2^3+3^3+4^3+5^3+6^3+7^3


Is there a general formula for S(k,a)=Sum(a^k)?



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Re: What is the Sum of 1^3+2^3+3^3+4^3+5^3+6^3+7^3 [#permalink] What is the Sum of 1^3+2^3+3^3+4^3+5^3+6^3+7^3   27 Jul 2013, 14:20
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alphabeta1234 wrote:
What is the sum of series S? Is there a shortcut/formula? I understand you can evaluate and sum each expression but I am trying to find a better solution.

S=1^3+2^3+3^3+4^3+5^3+6^3+7^3


Is there a general formula for S(k,a)=Sum(a^k)?


Hi,

i think there will be no need of these types of formulas in GMAT,BUT STILL HERE IT IS:

S=\(1^3+2^3+3^3+4^3+5^3+6^3+7^3+......+n^3\)\(=\) \((n(n+1)/2)^2\)


hope it helps
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Re: What is the Sum of 1^3+2^3+3^3+4^3+5^3+6^3+7^3 [#permalink] What is the Sum of 1^3+2^3+3^3+4^3+5^3+6^3+7^3   29 Jul 2013, 04:58
blueseas wrote:
alphabeta1234 wrote:
What is the sum of series S? Is there a shortcut/formula? I understand you can evaluate and sum each expression but I am trying to find a better solution.

S=1^3+2^3+3^3+4^3+5^3+6^3+7^3


Is there a general formula for S(k,a)=Sum(a^k)?


Hi,

i think there will be no need of these types of formulas in GMAT,BUT STILL HERE IT IS:

S=\(1^3+2^3+3^3+4^3+5^3+6^3+7^3+......+n^3\)\(=\) \((n(n+1)/2)^2\)

hope it helps


This was great!

How would you recommend doing this, though? Just by solving it out?
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Re: What is the Sum of 1^3+2^3+3^3+4^3+5^3+6^3+7^3 [#permalink] What is the Sum of 1^3+2^3+3^3+4^3+5^3+6^3+7^3   29 Jul 2013, 07:45
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sophietrophy wrote:
blueseas wrote:
alphabeta1234 wrote:
What is the sum of series S? Is there a shortcut/formula? I understand you can evaluate and sum each expression but I am trying to find a better solution.

S=1^3+2^3+3^3+4^3+5^3+6^3+7^3


Is there a general formula for S(k,a)=Sum(a^k)?


Hi,

i think there will be no need of these types of formulas in GMAT,BUT STILL HERE IT IS:

S=\(1^3+2^3+3^3+4^3+5^3+6^3+7^3+......+n^3\)\(=\) \((n(n+1)/2)^2\)

hope it helps


This was great!

How would you recommend doing this, though? Just by solving it out?


i would recommend to learn the formula in order to save time.

few more formulas are below.

 The sum of first n natural numbers = \(n(n+1)/2\)

 The sum of squares of first n natural numbers is \(n(n+1)(2n+1)/6\)

 The sum of cubes of first n natural numbers is\((n(n+1)/2)^2/4\)

 The sum of first n even numbers= \(n (n+1)\)

 The sum of first n odd numbers= \(n^2\)

hope it helps
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Re: What is the Sum of 1^3+2^3+3^3+4^3+5^3+6^3+7^3 [#permalink] What is the Sum of 1^3+2^3+3^3+4^3+5^3+6^3+7^3   30 Jul 2013, 00:13
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alphabeta1234 wrote:
What is the sum of series S? Is there a shortcut/formula? I understand you can evaluate and sum each expression but I am trying to find a better solution.

S=1^3+2^3+3^3+4^3+5^3+6^3+7^3


Is there a general formula for S(k,a)=Sum(a^k)?


You can solve it using pattern recognition. If you know the formula, it does save you time but then, there are many many formulas and you certainly cannot memorize all of them. So you should be comfortable with some such basic techniques. I have discussed how to use pattern recognition for this question here:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2013/01 ... n-part-ii/
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Re: What is the Sum of 1^3+2^3+3^3+4^3+5^3+6^3+7^3 [#permalink] What is the Sum of 1^3+2^3+3^3+4^3+5^3+6^3+7^3   03 Aug 2013, 08:26
What about when the formula starts with a number other than 1?

For example: S=4^3+5^3+6^3+7^3+...+18^3+19^3

How can we alter the formula [(n)(n+1)/2]^2 to account for this change?
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Re: What is the Sum of 1^3+2^3+3^3+4^3+5^3+6^3+7^3 [#permalink] What is the Sum of 1^3+2^3+3^3+4^3+5^3+6^3+7^3   03 Aug 2013, 08:33
alphabeta1234 wrote:
What about when the formula starts with a number other than 1?

For example: S=4^3+5^3+6^3+7^3+...+18^3+19^3

How can we alter the formula [(n)(n+1)/2]^2 to account for this change?


you cant change the formula...

but we can do in the following way:

suppose asked question:
S=4^3+5^3+6^3+7^3+...+18^3+19^3
= (1^3+2^3+3^3+4^3+5^3+6^3+7^3+...+18^3+19^3) - 1^3+2^3+3^3
= (19*20/2)^2 - (1+8+27)

hope it helps



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