Give an Inductive Proof that n^2 + n is even for all integers n >= 1
Thank you kindly.
Of course you don't need an inductive proof to show the result, but I gave an inductive proof since you asked for it. It is obvious that the expression is always even because it is the same as n(n+1) which, trivially, is always even for integer n.
It is true for n=1.
Assume it is true for n. So n^2 + n is even by assumption. Now look at it for n+1
Expand[(n^2 + n )/. n -> n + 1]
this is
2 + 3 n + n^2
which is the same as
(n^2 + n) + 2n + 2
Since is even, the above is even as well: QED.