# Sum of Sequence of Products of 3 Consecutive Integers

## 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:

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

$\blacksquare$