Question

Prove that the difference between two consecutive square numbers is always odd

Answer 1

The difference between two consecutive square numbers is always odd.

Step-by-step explanation:

To prove that the difference between two consecutive square numbers is always odd, we can use a simple mathematical proof. Let's consider two consecutive square numbers, n^2 and (n+1)^2. Taking the difference between them, we have:

(n+1)^2 - n^2 = n^2 + 2n + 1 - n^2 = 2n + 1.

We can see that the result is always of the form 2n+1, where n is an integer. Since an odd number is always of the form 2n+1, we can conclude that the difference between two consecutive square numbers is always odd.

Answer 2

✽ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ✽

➷ The first square number can be represented by
`n^(2)`

The second square number can be represented by
`(n+1)^(2)`

Subtract them:

`(n+1)^(2) - n^(2)`

Expand it first:

`n^(2) + 2n + 1 - n^(2)`

Simplify:

`2n+1`

Anything multiplied by 2 is even. Adding 1 to this value would make it odd. This proves it.

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

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