6. Prove that if 3n^3+13 is odd then n is even for all integers n, using the proof by contradiction.

Answer: If 3n^3+13 is odd then n is even for all integers n, using the proof by contradiction.Step-by-step explanation:By contradiction method, we need to prove that if [tex]3n^3+13[/tex] is odd then n is even for all integers n.Proof by contradiction:Let as assume if [tex]3n^3+13[/tex] is odd then n is odd for all integers n.[tex]n=2k+1[/tex]Substitute the value of n in the given expression.[tex]3(2k+1)^3+13[/tex]Cube of any odd number is an odd number.[tex](2k+1)^3=Odd[/tex]Product of two odd numbers is an odd number.[tex]3(2k+1)^3=Odd[/tex][tex]3(2k+1)^3=Odd[/tex] is an odd number and 13 is an odd number. We know that addition of two odd numbers is an even number.[tex]3(2k+1)^3+13=Even[/tex]Which is the contradiction of our assumption.If 3n^3+13 is odd then n is even for all integers n, using the proof by contradiction.Hence proved.