Question

Prove using the principle of mathematical induction $n \in {\text{N: }}n\left( {n + 1} \right)\left( {n + 5} \right)$ is a multiple of $3$.

Answer 1

Here, we will use the principle of mathematical induction which normally used to prove that any statement $P\left( n \right)$ holds for every natural number $n = 0,1,2,3,....$ , that is, the overall statement is a sequence of infinitely many cases $P\left( 0 \right)$ , $P\left( 1 \right)$ , $P\left( 2 \right)$ , …...

Formula used:
Complete step-by-step answer:
Step 1: We know that if a number is a multiple of $3$ , then it will come in the table $3$ as shown below:

$$3 \times 1 = 3 \\\ 3 \times 2 = 6 \\\ 3 \times 3 = 9 \\\ 3 \times 4 = 12$$

……..
$\Rightarrow$ Any number which is a multiple
$3$ will be equal to the product of any natural number $3$.
$\Rightarrow 3 = 3 \times {\text{N}}$ , where ${\text{N}}$ is the natural number.
Step 2: Suppose, $P\left( n \right):n\left( {n + 1} \right)\left( {n + 5} \right) = 3d$, where $d \in {\text{N}}$.
Now, by substituting the values of $n = 1$ in the equation $P\left( n \right):n\left( {n + 1} \right)\left( {n + 5} \right) = 3d$ , we get:
$\Rightarrow P\left( 1 \right):1\left( {1 + 1} \right)\left( {1 + 5} \right) = 3d$
By solving the LHS side:
$\Rightarrow {\text{L}}.{\text{H}}.{\text{S = }}1\left( 2 \right)\left( 6 \right)$
By doing multiplication, we get:
$\Rightarrow {\text{L}}.{\text{H}}.{\text{S = 12}}$
We can write $12$ as a multiple of $3$ as shown below:
$\Rightarrow {\text{L}}.{\text{H}}.{\text{S = 3}} \times \left( {\text{4}} \right)$
$\Rightarrow {\text{3}} \times \left( {\text{4}} \right) = 3d$, where $d = 4$.
$\Rightarrow {\text{L}}.{\text{H}}.{\text{S = R}}{\text{.H}}{\text{.S}}$
We can say that $P\left( n \right)$ is true for $n = 1$.
Step 3: Now we need to prove it for $n = k$ where $k \in {\text{N}}$.
Let’s assume that $P\left( k \right)$ is true:
Now we can write the above-given expression $P\left( n \right):n\left( {n + 1} \right)\left( {n + 5} \right) = 3d$ as below:
$\Rightarrow k\left( {k + 1} \right)\left( {k + 5} \right) = 3d$ where $k \in {\text{N}}$
By multiplying $k$ with $k + 1$ we get:
$\Rightarrow \left( {{k^2} + k} \right)\left( {k + 5} \right) = 3d$
Now by multiplying the brackets into the LHS side of the above equation, we get:
$\Rightarrow {k^2}\left( {k + 5} \right) + k\left( {k + 5} \right) = 3d$
$\Rightarrow {k^3} + 5{k^2} + {k^2} + 5k = 3d$
By adding the coefficients of ${k^2}$ , we get:
$\Rightarrow {k^3} + 6{k^2} + 5k = 3d$ ……………… (1)
Step 4: Now we will prove that $P\left( {k + 1} \right)$ is true:
By substituting the value of $n = k + 1$ in the equation $P\left( n \right):n\left( {n + 1} \right)\left( {n + 5} \right) = 3d$, we get:
$P\left( {k + 1} \right):\left( {k + 1} \right)\left( {\left( {k + 1} \right) + 1} \right)\left( {\left( {k + 1} \right) + 5} \right) = 3d$
By only solving LHS side:
$\Rightarrow {\text{L}}.{\text{H}}.{\text{S}} = \left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 6} \right)$
By doing multiplication of the first two brackets in the expression ${\text{L}}.{\text{H}}.{\text{S}} = \left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 6} \right)$, we get:
$\Rightarrow {\text{L}}.{\text{H}}.{\text{S}} = \left( {{k^2} + 2k + k + 2} \right)\left( {k + 6} \right)$
$\Rightarrow {\text{L}}.{\text{H}}.{\text{S}} = \left( {{k^2} + 3k + 2} \right)\left( {k + 6} \right)$
By doing multiplication of pending brackets, we get:
$\Rightarrow {\text{L}}.{\text{H}}.{\text{S}} = \left( {{k^3} + 3{k^2} + 2k + 6{k^2} + 18k + 12} \right)$
By adding coefficients of ${k^2}$ and $k$, we get:
$\Rightarrow {\text{L}}.{\text{H}}.{\text{S}} = \left( {{k^3} + 9{k^2} + 20k + 12} \right)$
Now from the equation (1), we will find the value of ${k^3}$ and substituting it in the expression ${\text{L}}.{\text{H}}.{\text{S}} = \left( {{k^3} + 9{k^2} + 20k + 12} \right)$ , we get:
$\Rightarrow {k^3} = 3d - 6{k^2} - 5k$ (from equation (1))
$\Rightarrow {\text{L}}.{\text{H}}.{\text{S}} = \left( {3d - 6{k^2} - 5k} \right) + 9{k^2} + 20k + 12$
By opening the brackets and doing simple addition and subtraction, we get:
$\Rightarrow {\text{L}}.{\text{H}}.{\text{S}} = 3d - 6{k^2} + 9{k^2} - 5k + 20k + 12$
$\Rightarrow {\text{L}}.{\text{H}}.{\text{S}} = 3d + 3{k^2} + 15k + 12$
Taking $3$ common from the above expression we get:
$\Rightarrow {\text{L}}.{\text{H}}.{\text{S}} = 3\left( {d + {k^2} + 5k + 4} \right)$
$\Rightarrow {\text{L}}.{\text{H}}.{\text{S}} = 3\left( m \right)$ , where $m = \left( {d + {k^2} + 5k + 4} \right)$ which is a natural number.
$\therefore P\left( {k + 1} \right)$is true whenever $P\left( k \right)$ is true.
From here, we can say that $P\left( n \right)$ is true for all-natural numbers.

Hence proved that $n \in {\text{N: }}n\left( {n + 1} \right)\left( {n + 5} \right)$ is a multiple of $3$.

Note: Students needs to remember the below principles of mathematical induction:
Let’s consider a given statement $P\left( n \right)$ involving natural number $n$ , so that:
The statement is true for $n = 1$ i.e. $P\left( 1 \right)$ is true, and
If the statement is true for $n = k$, where $k$ is a positive integer, then the statement is also true for all cases of $n = k + 1$, $P\left( k \right)$ leads to the truth of $P\left( {k + 1} \right)$.
This means that $P\left( n \right)$ is true for all the natural numbers, $n$.