Show 2 different solutions to the task.Prove that for every integer n (...-3, -2, -1, 0, 1, 2, 3, 4...), the expression n2 + n will always be even.

Answer with Step-by-step explanation:1. We are given that an expression [tex]n^2+n[/tex]We have to prove that this expression is always is even for every integer.There are two cases 1.n is odd integer 2.n is even integer1.n is an odd positive integer n square is also odd integer and n is odd .The sum of two odd integers is always even.When is negative odd integer then n square is positive odd integer and n is negative odd integer.We know that difference of two odd integers is always even integer.Therefore, given expression is always even .2.When n is even positive integer Then n square is always positive even integer and n is positive integer .The sum of two even integers is always even.Hence, given expression is always even when n is even positive integer.When n is negative even integer n square is always positive even integer and n is even negative integer .The difference of two even integers is always even integer.Hence, the given expression is always even for every integer.2.By mathematical inductionSuppose n=1 then n= substituting in the given expression1+1=2 =Even integerHence, it is true for n=1 Suppose it is true for n=kthen [tex]k^2+k[/tex] is even integer We shall prove that it is true for n=k+1[tex](k+1)^1+k+1[/tex]=[tex]k^1+2k+1+k+1[/tex]=[tex]k^2+k+2k+2[/tex]=Even +2(k+1)[/tex] because [tex]k^2+k[/tex] is even =Sum is even because sum even numbers is also even Hence, the given expression is always even for every integer n.