Sum of Sequence of Products of Consecutive Odd and Consecutive Even Numbers
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Theorem
\(\ds \forall n \in \Z_{>0}: \, \) | \(\ds \sum_{j \mathop = 1}^n j \paren {j + 2}\) | \(=\) | \(\ds 1 \times 3 + 2 \times 4 + 3 \times 5 + \dotsb + n \paren {n + 2}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {n \paren {n + 1} \paren {2 n + 7} } 6\) |
Proof
The proof proceeds by induction.
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
- $\ds \sum_{j \mathop = 1}^n j \paren {j + 2} = \frac {n \paren {n + 1} \paren {2 n + 7} } 6$
Basis for the Induction
$\map P 1$ is the case:
\(\ds \sum_{j \mathop = 1}^1 j \paren {j + 2}\) | \(=\) | \(\ds 1 \times 3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {1 \paren {1 + 1} \paren {2 \times 1 + 7} } 6\) |
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:
- $\ds \sum_{j \mathop = 1}^k j \paren {j + 2} = \frac {k \paren {k + 1} \paren {2 k + 7} } 6$
from which it is to be shown that:
- $\ds \sum_{j \mathop = 1}^{k + 1} j \paren {j + 2} = \frac {\paren {k + 1} \paren {k + 2} \paren {2 \paren {k + 1} + 7} } 6$
Induction Step
This is the induction step:
\(\ds \sum_{j \mathop = 1}^{k + 1} j \paren {j + 2}\) | \(=\) | \(\ds \sum_{j \mathop = 1}^k j \paren {j + 2} + \paren {k + 1} \paren {k + 3}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {k \paren {k + 1} \paren {2 k + 7} } 6 + \paren {k + 1} \paren {k + 3}\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {k + 1} \paren {\frac {k \paren {2 k + 7} + 6 \paren {k + 3} } 6}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {k + 1} \paren {\frac {2 k^2 + 13 k + 18} 6 }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {k + 1} \paren {k + 2} \paren {2 k + 9} } 6\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {k + 1} \paren {k + 2} \paren {2 \paren {k + 1} + 7} } 6\) |
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\ds \forall n \in \Z_{>0}: \sum_{j \mathop = 1}^n j \paren {j + 2} = \frac {n \paren {n + 1} \paren {2 n + 7} } 6$
$\blacksquare$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text I$. Algebra: The Method of Induction: Exercises $\text {II}$: $3$
- 1977: Gary Chartrand: Introductory Graph Theory ... (previous) ... (next): Appendix $\text{A}.6$: Mathematical Induction: Problem Set $\text{A}.6$: $40$