Q. 125.0( 2 Votes )

Prove by direct method that for any integer ‘n’, n3 – n is always even.

Answer :

We have given, n3-n

Let us Assume, n is even


Let n=2k, where k is natural number


n3-n=(2k)3-(2k)


n3-n=2k(4k2-1)


let k(4k2-1)=m


n3-n=2m


Therefore, (n3-n) is even.


Now, Let us Assume n is odd


Let n=(2k+1), where k is natural number


n3-n=(2k+1)3-(2k+1)


n3-n= (2k+1)[(2k+1)2-1]


n3-n= (2k+1)[(4k2+4k+1-1)]


n3-n= (2k+1)[(4k2+4k)]


n3-n= 4k(2k+1)(k+1)


n2-n= 2.2k(2k+1)(k+1)


let λ=2k(2k+1)(k+1)


n3-n=2λ


therefore, n3-n is even.


Hence, n3-n is always even


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