Question

Show that n(n+1)(n+2) is a multiple of 6 using mathematical induction

Answer 1

Answer:

Use principle of induction

Hypothesis n(n+1)(n+2) is a multiple of 6

when n=1

n(n+1)(n+2) =1(1+1)(1+2) =6 hence true for n=1

Assume it is true for n=k

i.e k(k+1)(k+2) is a multiple of

To prove that it is also true for n= k+1 if we assume it is true for n=k

For n=k+1

n(n+1)(n+2) =(k+1)(k+1+1)(k+1+2)

=(k+1)(k+2)(k+3) = (k+1)(k+2)(k) +(k+1)(k+2)(3)

=Multiple of 6 + (k+1)(k+2)(3)

For (k+1)(k+2)(3) if k is odd k+1 is even

Hence (k+1)(k+2)(3)is a multiple of 6

Whenk is eve k+2 is even (k+1)(k+2)(3) is multiple of 6

Hence n(n+1)(n+2) is a multiple of 6 for all values of n Є N

Answer 2

Step-by-step explanation:

The induction hypothesis - P(k):k(k+1)(2k+1) is divisible by 6, i.e. k(k+1)(2k+1)=6m for some m.

Now,

(k+1){(k+1)+1}{2(k+1)+1}

= (k+1)(k+2)(2k+3)

= k(k+1)(2k+3)+2(k+1)(2k+3)

= k(k+1)(2k+1)+2k(k+1)+2(k+1)(2k+3)

= 6m+(k+1)(2k+4k+6)

= 6m+6(k+1)2

So (k+1){(k+1)+1}{2(k+1)+1} is divisible by 6 i.e. P(k+1) is true.

Hope this helps you.

Thank You.