Question

Show 2 different solutions to the task.

Answer 1

Answer with Step-by-step explanation:

1. We are given that an expression $n^2+n$

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 integer

1.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 induction

Suppose n=1 then n= substituting in the given expression

1+1=2 =Even integer

Hence, it is true for n=1

Suppose it is true for n=k

then $k^2+k$ is even integer

We shall prove that it is true for n=k+1

$(k+1)^1+k+1$

=$k^1+2k+1+k+1$

=$k^2+k+2k+2$

=Even +2(k+1)[/tex] because $k^2+k$ is even

=Sum is even because sum even numbers is also even

Hence, the given expression is always even for every integer n.