Question

5. Prove that n^2-n is divisible by 2 for every positive integer'n' ?​

Answer 1

Answer:

Step-by-step explanation:

$n^{2} - n$

= n(n-1)

=(n)(n-1)                                                                          . . . .(i)

let n is even = 2q

$n(n-1) = 2q(2q-1)$

=2m                                       (where m is multiple of)

let n is odd = 2q +1

n(n-1) = (2q + 1) (2q + 1 -1)

=(2q-1) (2q)

=2m                                       (        "       )

hence $n^{2} -n$ is divisible by 1 for any positive integer  n

hope this helps

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