Question

prove by the principal of mathematical induction: n(n+1)(n+5) is a multiple of 3​

Answer 1

Proof:

An important term needed to be known in advance is 'Multiple'.

If one quantity is the mulitiple.of another quantity, then it shows that the if that particular quantity(the multiple) when divided by the source(from which the multiple came) gives an integer(except zero).

Now, nother thing regarding divisibility is that suppose if 'axbxc' is a multiple of 'd' then if by dividing 'axbxc' by 'd' should give an integer(no decimal point value).

However, if in the term 'axbxc', any factor i.e. either 'a' or 'b' or 'c' is divisble by 'd' will make the whole.term divisible.

To understand this, consider the following example:

If one needs check that if '3' is a factor of '100x105x45', then simply check individually. And we know that '45' form the above term is divisible by '3' and simultaneously will make the whole term divisible.

Now the actual proof.

We are given a term 'n(n+1)(n+5)' and we need to proof that it can be divided by '3' (multiple of '3').

So, for that, assuming that the given term is a multiple of '3' and we will proof for every possible.term, so obviously, at the end, our goal will be achieved. So firstly,simplifying the term a bit:

n(n+1)(n+5) = n³+6n²+5n = p(n)

where,

p(n) denotes a polynomial of 'n'.

Now, assume a kth term, so for its value, put 'k' in the polynomial;

p(k) = k³ + 6k² + 5k

Now for every next term i.e. '(k+1)th' term, again.put it in the polynomial the original polynomial:

p(k+1) = (k+1)(k+1+1)(k+1+5)

p(k+1) = (k+1)(k+2)(k+6)

p(k+1) = (k+1)(k²+8k+12)

p(k+1) = k³+8k²+12k+k²+8k+12

p(k+1) = k³+9k²+20k+12

p(k+1) = (k³+6k²+5k) +15k+12

p(k+1) = p(k) +15k +12

Now note that, p(k) is divisible by '3' as we have already mentioned and supposed, '15k' is also divisible as '15' is divisible, that will make '15k' entirely divisible and '12' is also divisible. Thus the entire term is divisible by '3'.