Prove that n(n+3) represents an even number for any natural number, n.?

Question

Can anyone help me with this math question/proof ?

Answer 1

Proof:
Case 1
Suppose n is odd, then n=2k+1, k is some integer. By substitution (2k+1)(2k+1+3)=(2k+1)(2k+4)= 4k^2+12k+4 which is even since 4k^2=2(2k^2)=2p= definition of even. basically, 4k^2+12k+4= 2(k^2+12k+4)=2t = definition of even
Similarly, Case 2
Suppose n is even, then n = 2m, m is some integer. By substitution 2m(2m+3)= 4m^2+6m. which is even
QED

Answer 2

well n(n+3) let n respresent a natural no so it equal to (n+3)=0 therefore n=-3 now if u subsitute -3 in the equation u get zero and zero is a even number if u subsitute n=1 which is a odd number u get 4 i think this is correct about 50%

Answer 3

The other guy is right but perhaps I can work it out for you:

if n is even that is:
n=2m where m is some integer
so n(n+3) is 2m(2m+3)
=4m^2 +6 m
=2(2m^2+3m) but (****) is just another integer
so if n is even then n(n+3) is even
and similar for the case when n is odd
but it seems trivial to me too..

Answer 4

Any natural number (n) must be odd or even. So n+3 will be the opposite. Multiply together an odd and an even number, and the product must be even (it is divisible by that even number that just produced it). This would work with (n+ any odd number).

Answer 5

This one is trivial. Clearly, either n is even or n+3 is even. The product of any number with an even number is even. QED.

Answer 6

1= -1 therefore your question is actually: prove that n(n-3) represents an eve………n.?

Answer 7

employing induction educate for the organic numbers n a million^2+2^2+3^2+...n^2=(n(n+a million)(2n+a million))/6 pf if n=a million then obviously a million^2 = (a million)(2)(3)/6 =a million so the reality is authentic even as n=a million. anticipate reality is authentic for n=ok. it fairly is we anticipate a million^2+2^2+3^2+...ok^2=(ok(ok+a million)(2k+a million))/6 Now, we educate the reality is authentic for n=ok+a million. a million^2+2^2+3^2+...+ok^2+(ok+a million)^2 by employing induction hypothesis all of us recognize that a million^2+2^2+3^2+...+ok^2 = (ok(ok+a million)(2k+a million))/6 if we upload (ok+a million)^2 to each and every side we get: a million^2+2^2+3^2+...+ok^2 + (ok+a million)^2 = (ok(ok+a million)(2k+a million))/6 + (ok+a million)^2 Now we simplify the right hand side. (ok(ok+a million)(2k+a million))/6 + (ok+a million)^2 = ( ok(ok+a million)(2k+a million))/6 + 6(ok+a million)^2/6= ( ok(ok+a million)(2k+a million) + 6(ok+a million)^2 ) /6= [ (ok+a million)(ok(2k+a million) + 6(ok+a million) ) ]/6= [(ok+a million)(2k^2 +7k +6) ]/6= [(ok+a million)(2k + 3)(ok + 2) ]/6= [(ok+a million)(2(ok+a million)+a million)((ok+a million)+a million)]/6 this shows the reality is authentic for n =ok+a million so the reality is authentic for all n by employing induction.