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July 23rd, 2017, 07:48 AM   #1
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Prove arbitrary sum of odd cubes:

Hello,

I am currently studying a textbook that asks me to use the fact that:

$\displaystyle 1^3 + 2^3 + ... + n^3 = (n(n+1) /2)^2$

in order to prove that:

$\displaystyle 1^3 + 3^3 + ... + (2n+1)^3 = (n+1)^2(2n^2+4n+1)$

The book says to do this WITHOUT using mathematical induction.

Help will be appreciated.
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July 23rd, 2017, 08:10 AM   #2
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Can you figure out the sum of cubes of the EVEN numbers first?

$$2^3 + 4^3 + .... + (2n)^3 = 2^3(1^3 + 2^3 + ... + n^3)$$

Go from there.
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July 23rd, 2017, 10:35 AM   #3
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Ah, thank you that put me on the right track! Should I post the proof?
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July 23rd, 2017, 11:18 AM   #4
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If you want us to check it, sure.
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July 23rd, 2017, 01:30 PM   #5
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I'll post to make sure and to help others with a similar problem:

If we combine the following 2 arbitrary sums like so:
$\displaystyle
S1 = 1^3 + 3^3 + 5^3 +...+(2n+1)^3
$

$\displaystyle + $

$\displaystyle
S2 = 2^3 + 4^3 + 6^3 +...+(2n)^3
$

We can see that term by term they will result in the following sum:

$\displaystyle
S3 = 1^3 + 2^3 +...+ (2n)^3 + (2n+1)^3
$

In order to find $\displaystyle S1$, we will do:

$\displaystyle S1 = S3 - S2$

for which the formulas are already known:

$\displaystyle S1 = (2n+1)(2n+2)/2)^2 - 2n^2(n+1)^2$


$\displaystyle S1 = (2n+1)^2(n+1)^2 - 2n^2(n+1)^2$


$\displaystyle S1= (n+1)^2((2n+1)^2 - 2n^2)$


$\displaystyle S1= (n+1)^2(4n^2+4n+1-2n^2)$


$\displaystyle S1 = 1^3 + 3^3 + 5^3 +...+ (2n+1)^3 = (n+1)^3(2n^2+4n +1)$
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Last edited by greg1313; July 23rd, 2017 at 05:19 PM.
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July 23rd, 2017, 02:34 PM   #6
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Mistake at the end that I can't edit for some reason.


It should be $\displaystyle S1 = (n+1)^2(2n^2 + 4n +1)$.

Also, treat the random '$\displaystyle a$'s as if they were '$\displaystyle n$'s.

Last edited by skipjack; July 23rd, 2017 at 07:58 PM.
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July 23rd, 2017, 05:19 PM   #7
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I've edited your post.
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