Prove that 3|n(2n²+1)?

Question

Prove that 3|n(2n²+1)

Prove that 3 is divider of n(2n²+1)

Answer 1

we use induction for n +ve integer
for n=1
n(2n²+1)=1(2(1²)+1))=3,which is divisble by 3
let us suppose that it is also true for n=k
i.e.3|k(2k²+1)
now consider
for n=k+1
n(2n²+1)=(k+1)(2(k+1)²+1)=k(2(k²+2k+1)+1)+(2(k+1)²+1)
=k(2k²+1)+k(4k+2)+(2k²+4k+2+1)
=k(2k²+1)+4k²+2k+2k²+4k+3
=k(2k²+1)+3(2k²+2k+1)
clearly R.H.S is divisible by 3
so
n(2n²+1) is also divisble by 3 for n =k+1 hence we conclude from induction that n(2n²+1)is divisible by 3 for all +ve integer n.similarly we can show for n-ve integers.That's all

Answer 2

Simple tabulation, modulo n, of the three possible cases will do it.

. . n . . . . . . 0 . . . 1 . . . 2
. . n². . . . . . 0 . . . 1 . . . 1
2n² + 1 . . . . 1 . . . 0 . . . 0

Therefore we see that for every value of n, either n is divisible by 3, or if not, then 2n²+1 is divisible by 3.

Answer 3

Prove by induction.
first show that it is true when n=1
then show that if n is true then n+1 is also true.

when n = 1; 1(2+1) is divisible by 3

(n+1)(2(n+1)^2+1)
=(n+1)(2n^2+4n+2+1)
=(n+1)(2n^2+4n+3)
=2n^3+4n^2+3n+2n^2+4n+3
=2n^3+6n^2 +7n + 3
=(2n^3+n) +3(2n^2+2n+1)
=n(2n^2+1)+3(2n^2+2n+1)

3(2n^2+2n+1) is divisible by 3

so n(2n^2+1) is divisible by 3 for all n

Answer 4

n(2n²+1) mod 3
≡ n(2n²-2) + 3n mod 3
≡ 2n(n+1)(n-1) mod 3
≡ 0
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Ideas: (n-1), n, and (n+1) are three consecutive integers. Therefore, one of them must contain 3 as a factor.