Prove by mathematical induction that n(n+1) (2n+1) is a multiple of 6 for all n∈N

When n=1, LHS=1(1+1)(2+)=2*3=6 (which is divisible by 6)

Similarly, when n=k, k(k+1)(2k+) is also divisible by 6 i.e. k(k+1)(2k+1) = 6m ------ (1)

Now when n=k+1, we have to prove that (k+1)(k+1+1)[2(k+1) +1} is a multiple of 6 i.e.

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

Now from (1) - (k+1) = [6m] / (k)(2k+1) --------- (2)

So now LHS = (k+1)(k+2)(2k+3) = { [6m] / (k)(2k+1) } * (k+2)(2k+3) which is divisible by 6 and hence a multiple of 6.

[ HENCE PROVED ]

  • -1
n(n+1) (2n+1) ,is divisible by 6

  • 0
n(n+1)(2n+1) is divisible by 6

  • 3
Show that n n + 1 to 2n + 1 is a multiple of 6 for every natural number n
  • -1
What are you looking for?