Sum of Sequence of Cubes/Proof by Induction

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Theorem

$\displaystyle \sum_{i=1}^n i^3 = \left({\sum_{i=1}^n i}\right)^2 = \frac{n^2 \left({n + 1}\right)^2} 4$

$\displaystyle \sum_{i=1}^n i = \frac {n \left({n + 1}\right)} 2$

So:

$\displaystyle \left({\sum_{i=1}^n i}\right)^2 = \frac{n^2 \left({n + 1}\right)^2} 4$

Next we use induction on $n$ to show that $\displaystyle \sum_{i=1}^n i^3 = \frac{n^2 \left({n + 1}\right)^2} 4$.

The base case holds since $\displaystyle 1^3 = \frac{1 \left({1 + 1}\right)^2} 4$.

Now we need to show that if it holds for $n$, then it holds for $n + 1$.

$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle \sum_{i=1}^{n+1} i^3$	$=$	$\displaystyle \sum_{i=1}^n i^3 + \left({n + 1}\right)^3$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle \frac{n^2 \left({n + 1}\right)^2} 4 + \left({n + 1}\right)^3$	$\displaystyle $	$\displaystyle $	$\displaystyle $	(by the induction hypothesis)
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle \frac{n^4 + 2 n^3 + n^2} 4 + \frac {4 n^3 + 12 n^2 + 12 n + 4} 4$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle \frac{n^4 + 6 n^3 + 13 n^2 + 12 n + 4} 4$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle \frac{\left({n + 1}\right)^2 \left({n + 2}\right)^2} 4$	$\displaystyle $	$\displaystyle $	$\displaystyle $

$\blacksquare$

\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \sum_{i=1}^{n+1} i^3\)	\(=\)	\(\displaystyle \sum_{i=1}^n i^3 + \left({n + 1}\right)^3\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac{n^2 \left({n + 1}\right)^2} 4 + \left({n + 1}\right)^3\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	(by the induction hypothesis)
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac{n^4 + 2 n^3 + n^2} 4 + \frac {4 n^3 + 12 n^2 + 12 n + 4} 4\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac{n^4 + 6 n^3 + 13 n^2 + 12 n + 4} 4\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac{\left({n + 1}\right)^2 \left({n + 2}\right)^2} 4\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)