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July 23rd, 2017, 07:48 AM  #1 
Member Joined: Apr 2017 From: Canada Posts: 32 Thanks: 2  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. 
July 23rd, 2017, 08:10 AM  #2 
Senior Member Joined: Oct 2009 Posts: 232 Thanks: 84 
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. 
July 23rd, 2017, 10:35 AM  #3 
Member Joined: Apr 2017 From: Canada Posts: 32 Thanks: 2 
Ah, thank you that put me on the right track! Should I post the proof?

July 23rd, 2017, 11:18 AM  #4 
Senior Member Joined: May 2016 From: USA Posts: 904 Thanks: 359 
If you want us to check it, sure.

July 23rd, 2017, 01:30 PM  #5 
Member Joined: Apr 2017 From: Canada Posts: 32 Thanks: 2 
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: $\displaystyle S1 = (2n+1)(2n+2)/2)^2  2n^2(n+1)^2$ $\displaystyle S1 = (2n+1)^2(n+1)^2  2n^2(n+1)^2$ $\displaystyle S1= (n+1)^2((2n+1)^2  2n^2)$ $\displaystyle S1= (n+1)^2(4n^2+4n+12n^2)$ $\displaystyle S1 = 1^3 + 3^3 + 5^3 +...+ (2n+1)^3 = (n+1)^3(2n^2+4n +1)$ Last edited by greg1313; July 23rd, 2017 at 05:19 PM. 
July 23rd, 2017, 02:34 PM  #6 
Member Joined: Apr 2017 From: Canada Posts: 32 Thanks: 2 
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. 
July 23rd, 2017, 05:19 PM  #7 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,745 Thanks: 1001 Math Focus: Elementary mathematics and beyond 
I've edited your post.


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arbitrary, cubes, odd, prove, sum 
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