Question

prove that every positive integer n, n^3+n is even​

Answer 1

Step-by-step explanation:

Let n be even:

n^3=even

n=even

So, n^3+n

=even+even

=even

Now , let n be an odd positive integer:

n^3=odd

n=odd

So , n^3+n

=odd+odd

=even.

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Answer 2

Step-by-step explanation:

let n be any positive integer and b= 2, then by Euclid's division lemma,

n = 2q+r

where r =0,1 (0≤r<b)

if n = 2q (r= 0)

n³+n =( 2q)³ + 2q

= 8q³ +2q

= 2(4q³+q)

since this is divisible by 2 , it is even

If n = 2q+1

n³+n = (2q+1)³ + 2q+1

= 8q³ + 1 + 12q² + 6q + 2q+1

= 8q³ + 12q² +6q+2

= 2(4q3+6q²+3q+1)

since this is divisible by 2 , it is even

so.. we have proven that , for any positive integer n, n³+n is always even