Why is the sum of the squares of two consecutive even numbers always a multiple of 4?

Two consecutive odd numbers can be denoted by 2n-1 and 2n + 1, so the product of two consecutive odd numbers plus one is (2n-1)(2n+1) +1 = 4n^2 -1 +1 = (2n)^2. Therefore, the number is always a multiple of 4 and a square of an even number.

How do you prove that the sum of the squares of any two consecutive odd integers is even?

Prove that the sum of the squares of any two consecutive integers is always an odd number. Start by breaking up the question into parts:’always an odd number’ . An even number is any number divisible 2 e.g. 2n, an odd number is always 1 higher or lower than an even number e.g. 2n+1 or something similar.

What is the sum of two consecutive even numbers?

The sum of any two consecutive numbers is always odd. Example, 4 + 5 = 9; –8 + (–7) = –15.

Is the sum of two consecutive numbers is always an odd number?

So, that means that the sum of the numbers will be: Sum = average \times number of consecutive numbers. This means the sum has an odd number as a factor. But if you add two consecutive numbers, the answer is always an odd number.

How do you prove that N 2 n is even?

Prove: If n is an even integer, then n2 is even. – If n is even, then n = 2k for some integer k. – n2 = (2k)2 = 4k2 – Therefore, n = 2(2k2), which is even.

Which of the following is the correct proof to show that the sum of two odd integers is even?

The sum of two odd integers is even. Proof: If m and n are odd integers then there exists integers a,b such that m = 2a+1 and n = 2b+1. m + n = 2a+1+2b+1 = 2(a+b+1).

How do you find two consecutive even numbers?

For any two consecutive even numbers, the difference is 2. For example, 6 and 8 are two consecutive even numbers, their difference = 8 – 6 = 2. If ‘n’ is an odd number, then the sum of ‘n’ consecutive numbers will be divisible by ‘n’.

How do you prove that the sum of two consecutive numbers is always odd?

The sum of these is n + n+1, which is 2n+1. 2n is even and +1 makes it odd. Therefore the sum of any two consecutive integers is odd. The answer in my textbook is “2n + (2n + 1) = 4n +1”.

How do you prove that two odd numbers are odd?

The product of two odd numbers is an odd number. Let m and k be any integers. This means that 2m+1 and 2k+1 are odd numbers. 2 ( 2mk + m + k ) + 1 which is an odd number.

What is the sum of the squares of any 2 even integers?

9 Prove algebraically that the sum of the squares of any 2 even positive integers is always a multiple of 4. (Total for question 9 is 2 marks) 10 Prove algebraically that the sum of the squares of any 2 odd positive integers is always even- + IkP/ f + / 4 n + 2 (Total for question 10 is 2 marks)

How do you sum the square of two consecutive numbers?

If you’re summing the square of two consecutive numbers (n and n+1), then you know that the two numbers (n and n+1) have a different parity: simply one is odd and the other is even; they cannot both be even or both be odd. Thus, let’s assume that [math]n [/math] is the even number and [math]n + 1 [/math] is the odd number.

Is the sum of two even numbers always even?

Yes, the sum of two even numbers is always even. So, x and y are mutiples of 2. x=2a and y=2b ; where a and b are any two integers. This means x+y is also multiple of 2. So x+y is an even number. Bonus – The sum of two odd numbers is also always even.

Is the square of an even number always a multiple of four?

By definition, n is an integer, because an even number should always be the product of two and some other integer. If we square 2n, we will get 2n*2n, or 4n^2. Therefore, the square of an even number will always be a multiple of four.