My Math Forum Prove arbitrary sum of odd cubes:

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 July 23rd, 2017, 06:48 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, 07:10 AM #2 Senior Member   Joined: Oct 2009 Posts: 439 Thanks: 147 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. Thanks from greg1313 and Antoniomathgini
 July 23rd, 2017, 09:35 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, 10:18 AM #4 Senior Member   Joined: May 2016 From: USA Posts: 1,116 Thanks: 462 If you want us to check it, sure.
 July 23rd, 2017, 12:30 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: $\displaystyle S1 = (2n+1)(2n+2)/2)^2 - 2n^2(n+1)^2$ $\displaystyle S1 = (2n+1)^2(n+1)^2 - 2n^2(n+1)^2$ $\displaystyle S1= (n+1)^2((2n+1)^2 - 2n^2)$ $\displaystyle S1= (n+1)^2(4n^2+4n+1-2n^2)$ $\displaystyle S1 = 1^3 + 3^3 + 5^3 +...+ (2n+1)^3 = (n+1)^3(2n^2+4n +1)$ Thanks from greg1313 Last edited by greg1313; July 23rd, 2017 at 04:19 PM.
 July 23rd, 2017, 01:34 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 06:58 PM.
 July 23rd, 2017, 04:19 PM #7 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,851 Thanks: 1075 Math Focus: Elementary mathematics and beyond I've edited your post. Thanks from Antoniomathgini

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