Question

How can we prove using mathematical induction that `n^2+n+1` is an odd number if n is a natural number?

Answer 1

See Explanation.
or try again yourself keeping in mind that : -
Any even number can be represented as 2t and any odd number can be represented as 2t+1 where t is also a natural number

Explanation:

Assuming you know the general algorithm of Principle of mathematical induction;

1 . Checking if the statement is true for `n=1`.
`1^2+1+1 = 3` which is odd

2 . Assuming statement is true for some natural number `k`.
i.e. `k^2+k+1` is odd.
i.e. `k^2 + k +1 = 2l+1` where `l` is another natural number.

[any even number can be represented as 2t and any odd number can be represented as 2t+1 where t is also a natural number]

3 . To prove that statement is true for natural number next to `k` i.e. `k+1`.

`implies` to prove that `(k+1)^2+(k+1)+1` is also odd.
`(k+1)^2+(k+1)+1 = k^2 + 1 + 2k + k +1 +1`
`= (k^2 + k +1) +2k +2`
`= (2l +1) +2(k+1)` (from point 2.)
`=2l +1 +2(k+1)`
`= 2(l+k+1) +1`
`= 2p+1` `-=` ODD (where p is a natural number)

now since all three, `l, k, 1` are natural numbers, their sum i.e. `l+k+1` is also a natural number and can be represented by another natural number.