View source for TADME2E 1.18

<b>Step 1:</b> Show that the statement holds for the basis case <math>n = 1</math><br>

:<math>
\begin{align}
\sum_{i=1}^1 i^3    & = (\sum_{i=1}^1)^2 \\
\Leftrightarrow 1^3 & = 1^2 \\
\Leftrightarrow 1   & = 1 \\
\end{align}
</math><br>

<b>Step 2:</b> Assume that <math>n</math> holds.

<b>Step 3:</b> Show that on the assumption that the summation is true for ''n'', it follows that it is true for ''n + 1''.

:<math>
\begin{align}
\sum_{i=1}^{n+1} i^3                         & = (\sum_{i=1}^{n+1} i)^2 \\
\Leftrightarrow \sum_{i=1}^{n} i^3 + (n+1)^3 & = (\sum_{i=1}^{n} i + (n+1))^2 \\
                                             & = (\sum_{i=1}^n i)^2 + 2(\sum_{i=1}^n i)(n+1) + (n+1)^2 \\
                                             & = (\sum_{i=1}^n i)^2 + 2(\frac{n(n+1)}{2})(n+1) + (n+1)^2 \\
                                             & = (\sum_{i=1}^n i)^2 + n(n+1)^2 + (n+1)^2 \\
                                             & = (\sum_{i=1}^n i)^2 + (n+1) \cdot (n+1)^2 \\
\Leftrightarrow \sum_{i=1}^{n} i^3 + (n+1)^3 & = (\sum_{i=1}^n i)^2 + (n+1)^3 \\
\Leftrightarrow \sum_{i=1}^{n} i^3           & = (\sum_{i=1}^n i)^2 \\
\end{align}
</math><br>
This is the case of ''n'' assumed to be true in step 2, so ''n + 1'' holds as well.


Above the arithmetic series was used: <math>\sum_{i=1}^n i = \frac{n(n+1)}{2}</math>

QED

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