Sum of Sequence of Products of Consecutive Odd and Consecutive Even Numbers

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Theorem

\(\, \displaystyle \forall n \in \Z_{>0}: \, \) \(\displaystyle \sum_{j \mathop = 1}^n j \paren {j + 2}\) \(=\) \(\displaystyle 1 \times 3 + 2 \times 4 + 3 \times 5 + \dotsb + n \paren {n + 2}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {n \paren {n + 1} \paren {2 n + 7} } 6\)


Proof

The proof proceeds by induction.

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

$\displaystyle \sum_{j \mathop = 1}^n j \paren {j + 2} = \frac {n \paren {n + 1} \paren {2 n + 7} } 6$


Basis for the Induction

$\map P 1$ is the case:

\(\displaystyle \sum_{j \mathop = 1}^1 j \paren {j + 2}\) \(=\) \(\displaystyle 1 \times 3\)
\(\displaystyle \) \(=\) \(\displaystyle 3\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {1 \paren {1 + 1} \paren {2 \times 1 + 7} } 6\)


Thus $\map P 1$ is seen to hold.


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown 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 the induction hypothesis:

$\displaystyle \sum_{j \mathop = 1}^k j \paren {j + 2} = \frac {k \paren {k + 1} \paren {2 k + 7} } 6$


from which it is to be shown that:

$\displaystyle \sum_{j \mathop = 1}^{k + 1} j \paren {j + 2} = \frac {\paren {k + 1} \paren {k + 2} \paren {2 \paren {k + 1} + 7} } 6$


Induction Step

This is the induction step:

\(\displaystyle \sum_{j \mathop = 1}^{k + 1} j \paren {j + 2}\) \(=\) \(\displaystyle \sum_{j \mathop = 1}^k j \paren {j + 2} + \paren {k + 1} \paren {k + 3}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {k \paren {k + 1} \paren {2 k + 7} } 6 + \paren {k + 1} \paren {k + 3}\) Induction Hypothesis
\(\displaystyle \) \(=\) \(\displaystyle \paren {k + 1} \paren {\frac {k \paren {2 k + 7} + 6 \paren {k + 3} } 6}\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {k + 1} \paren {\frac {2 k^2 + 13 k + 18} 6 }\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\paren {k + 1} \paren {k + 2} \paren {2 k + 9} } 6\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\paren {k + 1} \paren {k + 2} \paren {2 \paren {k + 1} + 7} } 6\)

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


Therefore:

$\displaystyle \forall n \in \Z_{>0}: \sum_{j \mathop = 1}^n j \paren {j + 2} = \frac {n \paren {n + 1} \paren {2 n + 7} } 6$

$\blacksquare$


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