Graphically, a point is a solution to a system of two inequalities if and only if the point
2 answers:
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Answer with Step-by-step explanation:
1. We are given that an expression
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 is even integer
We shall prove that it is true for n=k+1
=
=
=Even +2(k+1)[/tex] because is even
=Sum is even because sum even numbers is also even
Hence, the given expression is always even for every integer n.