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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: 912 Thanks: 354 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, 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: 1,310 Thanks: 552 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: $\displaystyle S1 = (2n+1)(2n+2)/2)^2 - 2n^2(n+1)^2$ $\displaystyle S1 = (2n+1)^2(n+1)^2 - 2n^2(n+1)^2$ $\displaystyle S1= (n+1)^2((2n+1)^2 - 2n^2)$ $\displaystyle S1= (n+1)^2(4n^2+4n+1-2n^2)$ $\displaystyle S1 = 1^3 + 3^3 + 5^3 +...+ (2n+1)^3 = (n+1)^3(2n^2+4n +1)$ Thanks from greg1313 Last edited by greg1313; July 23rd, 2017 at 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: 7,982 Thanks: 1166 Math Focus: Elementary mathematics and beyond I've edited your post. Thanks from Antoniomathgini Tags arbitrary, cubes, odd, prove, sum Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post shunya Elementary Math 3 October 26th, 2015 08:28 AM zak100 Geometry 0 February 25th, 2015 09:14 PM caters Probability and Statistics 3 May 9th, 2014 11:43 PM ruit Algebra 4 September 1st, 2008 07:32 PM andy Algebra 6 August 2nd, 2007 04:44 PM

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