Question

PLEASE SOLVE THIS QUESTION
PROVE THAT n**2-n IS DIVISIBLE BY 2 FOR EVERY POSITIVE INTEGER n.
PLEASE SOLVE THIS QUESTION PLEASE

Answer 1

Answer:-

n²-n

n(n-1)

∴n can be odd or even

i) if n is odd.                                            ii)If n is even

Then  (n-1) is even.                                  Then (n-1) is odd

n(n-1)=odd×even                                       n(n-1)=even×odd    

odd×even is always even.                       even ×odd is always even.

∴In both possibalities n is even. So, n is divisible by 2.

Answer 2

GIVEN:

★A dividend $n^{2}-n$

★A divisor as 2 .

TO PROVE:

★$n^{2}-n$ is divisible by 2 , for any positive integer n.

CONCEPT USED:

★We would use some proven results of Euclid Division lemma.

ANSWER:

By Euclid's division lemma we know that,

Every positive integer is the form of 2q (even) or 2q+1 (odd). for some positive integer q

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So, we have now two cases,

Case 1:

When n = 2q.

Then,

=$n^{2}-n$

= $n(n-1)$

On substituting n = 2q,

=$2q(2q-1)$

=$2(2q^{2}-q)$

=$2m$ (letting $2q^{2}-q=m$)

Hence, it is divisible by 2.

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Case 2:

When n = 2q+1

=$n^{2}-n$

=$n(n-1)$

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

=$2q(2q+1)$

=$2(q^{2}+2q)$

Hence it is also divisible by 2 .

Hence proved