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 September 25th, 2015, 11:54 PM #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, 07:31 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}

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