Using the principle of mathematical induction, prove each of the following

Solution:

To Prove:

$n \times(n+1) \times(n+2)$ is multiple of 6

Let us prove this question by principle of mathematical induction (PMI) for all natural numbers

$n \times(n+1) \times(n+2)$ is multiple of 6

Let $P(n): n \times(n+1) \times(n+2)$, which is multiple of 6

For $n=1 P(n)$ is true since $1 \times(1+1) \times(1+2)=6$, which is multiple of 6

Assume P(k) is true for some positive integer k , ie,

$=k \times(k+1) \times(k+2)=6 m$, where $m \in N \ldots(1)$

We will now prove that P(k + 1) is true whenever P( k ) is true

Consider

$=(k+1) \times((k+1)+1) \times((k+1)+2)$

$=(k+1) \times\{k+2\} \times\{(k+2)+1\}$

$=[(k+1) \times(k+2) \times(k+2)]+(k+1) \times(k+2)$

$=[k \times(k+1) \times(k+2)+2 \times(k+1) \times(k+2)]+(k+1) \times(k+2)$

$=[6 m+2 \times(k+1) \times(k+2)]+(k+1) \times(k+2)$

$=6 m+3 \times(k+1) \times(k+2)$

Now, (k + 1) & (k + 2) are consecutive integers, so their product is even

Then, (k + 1) × (k + 2) = 2×w (even)

Therefore,

$=6 m+3 \times[2 \times w]$

$=6 m+6 \times w$

$=6(m+w)$

= 6×q where q = (m + w) is some natural number

Therefore

$(\mathrm{k}+1) \times((\mathrm{k}+1)+1) \times((\mathrm{k}+1)+2)$ is multiple of 6

Therefore, P (k + 1) is true whenever P(k) is true.

By the principle of mathematical induction, P(n) is true for all natural numbers, ie, N

Hence proved