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Another summation example:
\begin{theorem}
For any $n \geq 0$, 
$\Sigma_{i=1}^n i^3 = (\Sigma_{i=1}^n i)^2$.
\end{theorem}

\begin{proof}
Let $P(n)$ be the proposition for $n$. \\
By induction.  \\
{\em BASIS STEP:} $P(0)$ -- both are $0$. \\
{\em INDUCTIVE STEP:} Assume $P(n)$, i.e., 
\[\Sigma_{i=1}^n i^3 = (\Sigma_{i=1}^n i)^2.\] \\
For $P(n+1)$, we must show: \
\[\Sigma_{i=1}^{n+1} i^3 = (\Sigma_{i=1}^{n+1} i)^2.\] \\
\\
LHS is just 
\[\Sigma_{i=1}^n i^3 + (n+1)^3,\]
which by inductive hypothesis is equal to: \\
\[(\Sigma_{i=1}^n i)^2 + (n+1)^3.\] \\
\\
RHS is just 
\[(\Sigma_{i=1}^n i + (n+1))^2.\] \\
which equals
\[(\Sigma_{i=1}^n i)^2 + 2 (n+1)(\Sigma_{i=1}^n i) + (n+1)^2.\] \\
\\
Are these two equal?  The first terms drop out, so it remains to show: \\
\[(n+1)^3 = 2 (n+1)(\Sigma_{i=1}^n i) + (n+1)^2.\] \\
Dividing out $n+1$, it's enough to show: \\
\[(n+1)^2 = 2 (\Sigma_{i=1}^n i) + (n+1).\] \\
Now it's useful to substitute in the closed form proved above for the
sum, so it remains to check: \\
\[(n+1)^2 = 2 ({n (n+1) \over 2}) + (n+1).\] \\
This reduces by easy algebra.
\end{proof}

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