Prove the following by using the principle of mathematical induction for all n ∈ N: n (n + 1) (n + 5) is a multiple of 3
Prove the following by using the principle of mathematical induction for all n ∈ N: n (n + 1) (n + 5) is a multiple of 3

We can compose the given assertion as

 

    \[P\text{ }\left( n \right):\text{ }\left( n\text{ }+\text{ }1 \right)\text{ }\left( n\text{ }+\text{ }5 \right)\]

, which is a different of

    \[3\]

In the event that

    \[n\text{ }=\text{ }1\]

we get

    \[\left( 1\text{ }+\text{ }1 \right)\text{ }\left( 1\text{ }+\text{ }5 \right)\text{ }=\text{ }12,\]

which is a various of

    \[3\]

Which is valid.

Think about

    \[P\text{ }\left( k \right)\]

be valid for some certain number

    \[k\]

    \[\left( k\text{ }+\text{ }1 \right)\text{ }\left( k\text{ }+\text{ }5 \right)\]

is a different of

    \[3\]

    \[\left( k\text{ }+\text{ }1 \right)\text{ }\left( k\text{ }+\text{ }5 \right)\text{ }=\text{ }3m,\]

where

    \[m\in N\ldots \left( 1 \right)\]

Presently let us demonstrate that

    \[P\text{ }\left( k\text{ }+\text{ }1 \right)\]

is valid.

Here

    \[\left( k\text{ }+\text{ }1 \right)\text{ }\left\{ \left( k\text{ }+\text{ }1 \right)\text{ }+\text{ }1 \right\}\text{ }\left\{ \left( k\text{ }+\text{ }1 \right)\text{ }+\text{ }5 \right\}\]

We can compose it as

    \[=\text{ }\left( k\text{ }+\text{ }1 \right)\text{ }\left( k\text{ }+\text{ }2 \right)\text{ }\left\{ \left( k\text{ }+\text{ }5 \right)\text{ }+\text{ }1 \right\}\]

By duplicating the terms

    \[=\text{ }\left( k\text{ }+\text{ }1 \right)\text{ }\left( k\text{ }+\text{ }2 \right)\text{ }\left( k\text{ }+\text{ }5 \right)\text{ }+\text{ }\left( k\text{ }+\text{ }1 \right)\text{ }\left( k\text{ }+\text{ }2 \right)\]

So we get

    \[=\text{ }\left\{ k\text{ }\left( k\text{ }+\text{ }1 \right)\text{ }\left( k\text{ }+\text{ }5 \right)\text{ }+\text{ }2\text{ }\left( k\text{ }+\text{ }1 \right)\text{ }\left( k\text{ }+\text{ }5 \right) \right\}\text{ }+\text{ }\left( k\text{ }+\text{ }1 \right)\text{ }\left( k\text{ }+\text{ }2 \right)\]

Subbing condition

    \[\left( 1 \right)\]

    \[=\text{ }3m\text{ }+\text{ }\left( k\text{ }+\text{ }1 \right)\text{ }\left\{ 2\text{ }\left( k\text{ }+\text{ }5 \right)\text{ }+\text{ }\left( k\text{ }+\text{ }2 \right) \right\}\]

By increase

    \[=\text{ }3m\text{ }+\text{ }\left( k\text{ }+\text{ }1 \right)\text{ }\left\{ 2k\text{ }+\text{ }10\text{ }+\text{ }k\text{ }+\text{ }2 \right\}\]

On additional computation

    \[=\text{ }3m\text{ }+\text{ }\left( k\text{ }+\text{ }1 \right)\text{ }\left( 3k\text{ }+\text{ }12 \right)\]

Accepting

    \[3\]

as normal

    \[=\text{ }3m\text{ }+\text{ }3\text{ }\left( k\text{ }+\text{ }1 \right)\text{ }\left( k\text{ }+\text{ }4 \right)\]

We get

    \[<span class="ql-right-eqno"> (1) </span><span class="ql-left-eqno">   </span><img src="https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4e13cd15f09d16abc2754c01c26744c4_l3.png" height="80" width="705" class="ql-img-displayed-equation quicklatex-auto-format" alt="\begin{align*} & =\text{ }3\text{ }\left\{ m\text{ }+\text{ }\left( k\text{ }+\text{ }1 \right)\text{ }\left( k\text{ }+\text{ }4 \right) \right\} \\ & =\text{ }3\text{ }\times \text{ }q\text{ }where\text{ }q\text{ }=\text{ }\left\{ m\text{ }+\text{ }\left( k\text{ }+\text{ }1 \right)\text{ }\left( k\text{ }+\text{ }4 \right) \right\} \\ \end{align*}" title="Rendered by QuickLaTeX.com"/>\]

is some regular number

    \[\left( k\text{ }+\text{ }1 \right)\text{ }\left\{ \left( k\text{ }+\text{ }1 \right)\text{ }+\text{ }1 \right\}\text{ }\left\{ \left( k\text{ }+\text{ }1 \right)\text{ }+\text{ }5 \right\}\]

is a numerous of

    \[3\]

    \[P\text{ }\left( k\text{ }+\text{ }1 \right)\]

is valid at whatever point

    \[P\text{ }\left( k \right)\]

is valid.

Hence, by the rule of numerical enlistment, articulation

    \[P\text{ }\left( n \right)\]

is valid for all regular numbers for example

    \[n.\]