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What is the sum of the following sequence: 1000^3 - 999^3 - 998^3 - 997^3 -.............-3^3 - 2^3 - 1^3. Any hint or help would be greatly appreciated. Thank you.

 Mar 26, 2019
 #1
avatar+28288 
+4

Hint:

 

You can write this as \(1000^3-\sum_{k=1}^{999}k^3\)

 

Now \(\sum_{k=1}^nk^3=\frac{n^2(n+1)^2}{4}\)

 

Hope this helps!

 Mar 26, 2019
 #2
avatar
+1

Hello Alan: The second part evaluates to this:
∑[ (k^3), k, 1, 999] = 249,500,250,000, which is not correct. It should be written as follows:
∑[ (k - 1)^3, k, 999, 1] = 0, since the terms are continuously declining by 1.  

 Mar 26, 2019
edited by Guest  Mar 26, 2019
 #3
avatar+28288 
+3

I don't see it!  As far as I can see the negative terms run from 1 cubed to 999 cubed so this term is as I wrote it.

Alan  Mar 26, 2019
 #6
avatar+1821 
+2

The formal definition for summations defined recursively is

 

\(\sum \limits_{i=a}^{b}g(i)=0 \qquad | \text {for b < a}\\ \sum \limits_{i=a}^{b}g(i)=g(b)+\sum\limits _{i=a}^{b-1}g(i) \qquad| \text {for b > a }\\\)

 

Note that the zero (0) result is true for all summations where the lower limit is larger than the upper limit.

 

So, Mr. BB, wouldn’t it be rather pointless to present a finite series only to define it as zero (0)?  I suppose there may be reasons for this, but this series isn’t one of them.

 

GA

GingerAle  Mar 27, 2019
 #4
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+1

NO! Alan is correct!

 Mar 26, 2019
edited by Guest  Mar 26, 2019
edited by Guest  Mar 26, 2019
 #5
avatar+23575 
+2

What is the sum of the following sequence: 1000^3 - 999^3 - 998^3 - 997^3 -.............-3^3 - 2^3 - 1^3.

 

\(\begin{array}{|rcll|} \hline && \mathbf{ 1000^3 - 999^3 - 998^3 - 997^3 -\ldots -3^3 - 2^3 - 1^3} \\ \\ &=& 1000^3 - \left(1^3+2^3+3^3+\ldots +997^3+998^3+999^3 \right) \\ &=& 1000^3 - \left(1 +2 +3 +\ldots +997 +998 +999 \right)^2 \\ &=& 1000^3 - \left(\left(\dfrac{1+999}{2}\right) \cdot 999\right)^2 \\ &=& 1000^3 - 500^2 \cdot 999^2 \\ &=& 1\ 000\ 000\ 000 - 249\ 500\ 250\ 000 \\ &\mathbf{=}& \mathbf{-248\ 500\ 250\ 000} \\ \hline \end{array}\)

 

laugh

 Mar 27, 2019

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