Sum of Sequence of Products of Consecutive Integers

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Theorem

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

For all $n \in \N^*$, let $P \left({n}\right)$ be the proposition:

$\displaystyle \forall n \ge 1: \sum_{j = 1}^n j \left({j+1}\right) = \frac {n \left({n+1}\right) \left({n+2}\right)} 3$

$P(1)$ is true, as this just says $1 \times 2 = 2 = \dfrac {1 \times 2 \times 3} 3$.

Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.

$\displaystyle \sum_{j = 1}^k j \left({j+1}\right) = \frac {k \left({k+1}\right) \left({k+2}\right)} 3$

Then we need to show:

$\displaystyle \sum_{j = 1}^{k+1} j \left({j+1}\right) = \frac {\left({k+1}\right) \left({k+2}\right) \left({k+3}\right)} 3$

$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle \sum_{j = 1}^{k+1} j \left({j+1}\right)$	$=$	$\displaystyle \sum_{j = 1}^k j \left({j+1}\right) + \left({k+1}\right) \left({k+2}\right)$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle \frac {k \left({k+1}\right) \left({k+2}\right)} 3 + \left({k+1}\right) \left({k+2}\right)$	$\displaystyle $	$\displaystyle $	$\displaystyle $	Induction hypothesis
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle \frac {k \left({k+1}\right) \left({k+2}\right) + 3 \left({k+1}\right) \left({k+2}\right)} 3$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle \frac {\left({k+1}\right) \left({k+2}\right) \left({k+3}\right)} 3$	$\displaystyle $	$\displaystyle $	$\displaystyle $

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:

$\displaystyle \forall n \ge 1: \sum_{j = 1}^n j \left({j+1}\right) = \frac {n \left({n+1}\right) \left({n+2}\right)} 3$

$\blacksquare$

\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \sum_{j = 1}^{k+1} j \left({j+1}\right)\)	\(=\)	\(\displaystyle \sum_{j = 1}^k j \left({j+1}\right) + \left({k+1}\right) \left({k+2}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {k \left({k+1}\right) \left({k+2}\right)} 3 + \left({k+1}\right) \left({k+2}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Induction hypothesis
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {k \left({k+1}\right) \left({k+2}\right) + 3 \left({k+1}\right) \left({k+2}\right)} 3\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {\left({k+1}\right) \left({k+2}\right) \left({k+3}\right)} 3\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)