Question

By using the principle of mathematical induction,Prove that \[n(n + 1)(n + 5)\] is a multiple of 3.

Answer 1

Hint:In this question we have to prove that the \[n(n + 1)(n + 5)\] is a multiple of 3. Let assume \[P(n) = n(n + 1)(n + 5)\]. Hence we have to prove that \[P(n) = n(n + 1)(n + 5) = 3d\] where \[d \in N\] by using the principle of mathematical induction.

Complete step-by-step answer:
We have to prove that \[n(n + 1)(n + 5)\] is a multiple of 3
We will prove it by using the concept of mathematical induction method for all \[n \in N\]
Let \[P(n) = n(n + 1)(n + 5) = 3d\] where \[d \in N\]
For \[n = 1\]
Since \[P(n) = n(n + 1)(n + 5)\]
Substituting the value \[n = 1\]in above equation,
\[P(1) = 1(1 + 1)(1 + 5)\]
\[P(1) = 1(2)(6) = 12\] which is divisible by 3
\[P(n)\] is true for \[n = 1\]
By using the induction method
So, \[P(k)\] is also true
Again, we check for \[P(k)\]
We can write
\[P(k) = k(k + 1)(k + 5) = 3m\] where \[m \in N\]
Simplify and we get
\[ \Rightarrow {k^3} + 6{k^2} + 5k = 3m\]
\[ \Rightarrow {k^3} = - 6{k^2} - 5k + 3m\]
Now we will prove that \[P(k + 1)\] is true
so in the place of \[k\] replace \[k + 1\]
\[P(k + 1) = (k + 1)(k + 2)(k + 6) = {k^3} + 9{k^2} + 20k + 12\;\]
Putting the value of \[{k^3}\] in above equation we get,
\[ = (3m - 6{k^2} - 5k) + 9{k^2} + 20k + 12\]
Adding the terms have same power we get,
\[ = 3m + 3{k^2} + 15k + 12\]
Taking 3 as a common term,
\[ = 3(m + {k^2} + 5k + 4)\]
We can write \[3r\], where \[r = m + {k^2} + 5k + 4\]
We can see that \[P(k + 1)\]is multiply by 3
So \[P(k + 1)\] is also true.
Since \[P(k)\] is true whenever \[P(k)\] is true.
Hence proved \[n(n + 1)(n + 5)\] is a multiple of 3
So, by the principle of induction, \[P(n)\] is divisible by 3 for all \[n \in N\]
When a statement is true for a natural number \[n = k\], then it will also be true for its successor, \[n = k + 1\], and the statement is true for \[n = 1\], then the statement will be true for every natural number \[n\].

Note:Mathematical induction: A proof by induction consists of two cases. The first, the base case (or basis), proves the statement for \[n = \;1\] without assuming any knowledge of other cases. The second case, the induction step, proves that if the statement holds for any given case \[n = k\], then it must also hold for the next case \[n\; = \;k\; + {\text{ }}1\]. These two steps establish that the statement holds for every natural number \[n\].