Sum of Sequence of Products of Consecutive Integers/Proof 1

From ProofWiki
Jump to navigation Jump to search

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

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:

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

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Sum_of_Sequence_of_Products_of_Consecutive_Integers/Proof_1&oldid=596808"