maths problem HELPPP?

Question

its from mathematical induction
use the principle of mathematic induction to prove the following statewment for all natural numbers:
n(n+1)(n+5) is a multiple of 6

can u plz explain this step
(k+1)(k+2)(k+6)
= k(k+1)(k+5) + 3k^2 + 15k + 12

Answer 1

P(k) = k(k+1)(k+5) is multiple of 6, where k = 1,2,3...
P(1) = (1)(2)(6) = 12 (a multiple of 6)

P(k+1) = (k+1)(k+2)(k+6)
= (k+1)[k^2+8k+12]
= (k+1)[k^2+5k+3k+12]
= (k+1)(k^2+5k) + (k+1)(3k+12)
= (k+1)(k)(k+5) + 3(k+1)(k+4)
= P(k) + 3(k+1)(k+4)

Prove (k+1)(k+4) is multiple of 2 because 3 multiply an even number will be a multiple of 6.

So if k = even, (k+4) is even number.
if k = odd, (k+1) is even number.

Finally, P(k+1) will be multiple of 6.

Alternatively, you can assume Q(k) = (k+1)(k+4), find Q(k+1) = Q(k) + 2(k+6)

Answer 2

Step 1:
check that it is true for n = 1
Step 2:
assume the hypothesis to be true for n = k, i.e.

k(k+1)(k+5) is a multiple of 6

Step 3:

u have to prove the hypothesis to be true for n = k+1
i.e.
(k+1)(k+2)(k+6) is a multiple of 6

Proof:
(k+1)(k+2)(k+6)
= k(k+1)(k+5) + 3k^2 + 15k + 12

the first term is a multiple of 6 by step 2
the remaining part:
3k^2 + 15k + 12
= 3k^2 + 12k + 3k + 12
= 3k(k+4) + 3(k+4)
= 3(k+1)(k+4)

now all u need to do is to prove (k+1)(k+4) is a multiple of 2
which is obvious, because:

if k is even, then k+4 is even
if k is odd, then k+1 is even
so in either case (k+1)(k+4) is a multiple of 2

thus the hypothesis is true for n = k+1

so by principle of mathematical induction the given statement is true for all n

Answer 3

It is clear without mathematical induction. From 3 consecutive numbers surely one is divisible by 3 and at least one is divisiblle by 2.

Answer 4

well maths is a problem for me tooo so try to maintain it

Answer 5

substituting for n=k+1 u get those solution....................................