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Not sure where to begin with this problem.

Posted 10 years ago

Give an Inductive Proof that n^2 + n is even for all integers n >= 1

POSTED BY: Hektik x
3 Replies
Posted 10 years ago

Thank you kindly.

POSTED BY: Hektik x

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.

POSTED BY: David Reiss

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.

POSTED BY: David Reiss
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