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September 26th, 2015, 12:54 AM   #1
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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.
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September 26th, 2015, 08:31 AM   #2
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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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