# Sum of Sequence of Products of Consecutive Integers

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## Theorem

 $\ds \sum_{j \mathop = 1}^n j \paren {j + 1}$ $=$ $\ds 1 \times 2 + 2 \times 3 + \dotsb + n \paren {n + 1}$ $\ds$ $=$ $\ds \frac {n \paren {n + 1} \paren {n + 2} } 3$

## Proof 1

Proof by induction:

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

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

### Basis for the Induction

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

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:

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

Then we need to show:

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

### Induction Step

This is our induction step:

 $\ds \sum_{j \mathop = 1}^{k + 1} j \paren {j + 1}$ $=$ $\ds \sum_{j \mathop = 1}^k j \paren {j + 1} + \paren {k + 1} \paren {k + 2}$ $\ds$ $=$ $\ds \frac {k \paren {k + 1} \paren {k + 2} } 3 + \paren {k + 1} \paren {k + 2}$ Induction hypothesis $\ds$ $=$ $\ds \frac {k \paren {k + 1} \paren {k + 2} + 3 \paren {k + 1} \paren {k + 2} } 3$ $\ds$ $=$ $\ds \frac {\paren {k + 1} \paren {k + 2} \paren {k + 3} } 3$

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

Therefore:

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

$\blacksquare$

## Proof 2

Observe that:

 $\ds 3 i \paren {i + 1}$ $=$ $\ds i \paren {i + 1} \paren {i + 2} - i \paren {i + 1} \paren {i - 1}$ $\ds$ $=$ $\ds \paren {i + 1} \paren {\paren {i + 1} + 1} \paren {\paren {i + 1} - 1} - i \paren {i + 1} \paren {i - 1}$

Thus:

$3 i \paren {i + 1} = b \paren {i + 1} - b \paren i$

where:

$b \paren i = i \paren {i + 1} \paren {i - 1}$

This can therefore be used as the basis of a telescoping series, as follows:

 $\ds \sum_{i \mathop = 1}^n 3 i \paren {i + 1}$ $=$ $\ds \sum_{i \mathop = 1}^n \paren {i \paren {i + 1} \paren {i + 2} - i \paren {i + 1} \paren {i - 1} }$ $\ds$ $=$ $\ds n \paren {n + 1} \paren {n + 2} - 1 \paren {1 + 1} \paren {1 - 1}$ Telescoping Series $\ds$ $=$ $\ds n \paren {n + 1} \paren {n + 2}$

Hence the result.

$\blacksquare$

## Proof 3

 $\ds \sum_{j \mathop = 1}^n j \paren {j + 1}$ $=$ $\ds \sum_{j \mathop = 1}^n \paren {j^2 + j}$ $\ds$ $=$ $\ds \sum_{j \mathop = 1}^n j^2 + \sum_{j \mathop = 1}^n j$ $\ds$ $=$ $\ds \sum_{j \mathop = 1}^n j^2 + \frac {n \paren {n + 1} } 2$ Closed Form for Triangular Numbers $\ds$ $=$ $\ds \frac {n \paren {n + 1} \paren {2 n + 1} } 6 + \frac {n \paren {n + 1} } 2$ Sum of Sequence of Squares $\ds$ $=$ $\ds n \paren {n + 1} \paren {\frac {\paren {2 n + 1} } 6 + \frac 1 2}$ extracting $n \paren {n + 1}$ as a common factor $\ds$ $=$ $\ds \frac {n \paren {n + 1} \paren {n + 2} } 3$ simplifying