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:

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


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:

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


Then we need to show:

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


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:

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

$\blacksquare$


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