Name: Mathematical Induction: Illustration
Uploaded: 2019-07-19
Duration: 4 min 19 s
Description: The formula for the sum of first “n” natural numbers can be derived using the formula of the sum to n terms of AP. Could you think of any other method to der...

Question

Using the principle of mathematical induction, prove the following for $n \in N$ ;

$n (n + 1) (n + 2)$ is a multiple of $6$ .

Accepted Answer

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.

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 $m \in N . . . (1)$

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, $(k + 1) \times (k + 2) = 2 \times w$ (even)

Therefore,

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

$= 6 m + 6 \times w$

$= 6 (m + w)$

$= 6 \times q$ , where $q = (m + w)$ is some natural number

Therefore $(k + 1) \times ((k + 1) + 1) \times ((k + 1) + 2)$ 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.

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