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 September 26th, 2015, 12:54 AM #1 Senior Member   Joined: Dec 2014 From: The Asymptote Posts: 142 Thanks: 6 Math Focus: Certainty Telescoping Given the following sum, $\displaystyle 2^3 + 4^3 + 6^3 +...+ n^3$ A formula for the above is: $\displaystyle \sum^n_{i = 1} 2i^3$ I'm unfamiliar with the method of telescoping and have just stated the formula based on what I see, how does one deduce the formula using the method of telescoping?? Thank you in advance. September 26th, 2015, 08:31 AM #2 Member   Joined: Apr 2015 From: USA Posts: 46 Thanks: 32 Does telescoping include discrete "integration" where only the first and last (n+1) terms matter and all the terms in between cancel? \begin{align}\sum_{i=1}^{n}(2i)^3&=\sum_{i=1}^{n} 2^3i^3\\ &=2^3\sum_{i=1}^{n}i^3\quad\text{Now, express }i^3\text{ in terms of falling powers.}\\ &=8\sum_{i=1}^{n}i^{\underline{3}}+3 i^{\underline{2}}+i^{\underline{1}}\quad \text{"Integrate" using the (falling) power rule.}\\ &=8\left.\left(\frac{1}{4}i^{\underline{4}}+ i^{\underline{3}}+\frac{1}{2} i^{\underline{2}}\right)\right|_{i=1}^{n+1}\\ &=8\left.\left(\frac{1}{4}(i)(i-1)(i-2)(i-3)+(i)(i-1)(i-2)+\frac{1}{2}(i)(i-1)\right)\right|_{i=1}^{n+1}\\ &=8\left.\left(\frac{i^2(i-1)^2}{4}\right)\right|_{i=1}^{n+1}\\ &=8\left(\frac{(n+1)^2(n+1-1)^2}{4}-\frac{(1)^2\cancelto{0}{(1-1)^2}}{4}\right)\\ &=8\left(\frac{(n+1)^2\,n^2}{4}\right)\\ &=2(n+1)^2\,n^2\\ \end{align} Tags telescoping 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 rnck Pre-Calculus 5 October 27th, 2014 04:54 PM Raptor Calculus 4 May 7th, 2014 10:18 PM jiasyuen Math Events 2 December 21st, 2013 04:38 AM proglote Algebra 9 August 24th, 2011 04:35 PM truck_driver Calculus 7 April 27th, 2008 02:09 PM

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