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 |
Sources
- 1971: George E. Andrews: Number Theory ... (previous) ... (next): $\text {1-1}$ Principle of Mathematical Induction: Exercise $4$
- 1971: Robert H. Kasriel: Undergraduate Topology ... (previous) ... (next): $\S 1.17$: Finite Induction and Well-Ordering for Positive Integers: Exercise $6$