Prove that (n²-n) is even for every positive integer n ?
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(n^2-n)
=n(n-1)
here n is even and n-1 is odd.
since we know that even*odd is always even, thus (n^2-n) is always even
=n(n-1)
here n is even and n-1 is odd.
since we know that even*odd is always even, thus (n^2-n) is always even
AkanshSaini:
Thank you @srijan007
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