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

Hint: 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, \[n\].