Sum of Sequence of Products of 3 Consecutive Integers

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Theorem

\(\ds \sum_{j \mathop = 1}^n j \paren {j + 1} \paren {j + 2}\) \(=\) \(\ds 1 \times 2 \times 3 + 2 \times 3 \times 4 + \cdots + \paren {n - 1} n \paren {n + 1} + n \paren {n + 1} \paren {n + 2}\)
\(\ds \) \(=\) \(\ds \dfrac {n \paren {n + 1} \paren {n + 2} \paren {n + 3} } 4\)

Proof

Proof by induction:

For all $n \in \N_{>0}$, let $\map P n$ be the proposition:

$\ds \forall n \ge 1: \sum_{j \mathop = 1}^n j \paren {j + 1} \paren {j + 2} = \dfrac {n \paren {n + 1} \paren {n + 2} \paren {n + 3} } 4$


Basis for the Induction

$\map P 1$ is the case:

\(\ds \sum_{j \mathop = 1}^1 j \paren {j + 1} \paren {j + 2}\) \(=\) \(\ds 1 \times 2 \times 3\)
\(\ds \) \(=\) \(\ds \dfrac {1 \times 2 \times 3 \times 4} 4\)
\(\ds \) \(=\) \(\ds \dfrac {1 \paren {1 + 1} \paren {1 + 2} \paren {1 + 3} } 4\)

Thus $\map P 1$ is seen to be true.

This is our basis for the induction.


Induction Hypothesis

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


So this is our induction hypothesis:

$\ds \sum_{j \mathop = 1}^k j \paren {j + 1} \paren {j + 2} = \dfrac {k \paren {k + 1} \paren {k + 2} \paren {k + 3} } 4$


Then we need to show:

$\ds \sum_{j \mathop = 1}^{k + 1} j \paren {j + 1} \paren {j + 2} = \dfrac {\paren {k + 1} \paren {k + 2} \paren {k + 3} \paren {k + 4} } 4$


Induction Step

This is our induction step:

\(\ds \sum_{j \mathop = 1}^{k + 1} j \paren {j + 1} \paren {j + 2}\) \(=\) \(\ds \sum_{j \mathop = 1}^k j \paren {j + 1} \paren {j + 2} + \paren {k + 1} \paren {k + 2} \paren {k + 3}\)
\(\ds \) \(=\) \(\ds \dfrac {k \paren {k + 1} \paren {k + 2} \paren {k + 3} } 4 + \paren {k + 1} \paren {k + 2} \paren {k + 3}\) Induction hypothesis
\(\ds \) \(=\) \(\ds \dfrac {k \paren {k + 1} \paren {k + 2} \paren {k + 3} + 4 \paren {k + 1} \paren {k + 2} \paren {k + 3} } 4\)
\(\ds \) \(=\) \(\ds \dfrac {\paren {k + 1} \paren {k + 2} \paren {k + 3} \paren {k + 4} } 4\) algebra

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


Therefore:

$\ds \forall n \ge 1: \sum_{j \mathop = 1}^n j \paren {j + 1} \paren {j + 2} = \dfrac {n \paren {n + 1} \paren {n + 2} \paren {n + 3} } 4$

$\blacksquare$


Sources

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