The product of two consecutive positive integers is added to the larger of the two integers. Prove that the result is always a square number.

think: (a+1)^2 = a^2+2a+1

n (n+1) + n+1

n^2 + n + n + 1

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

Thank you so much.

3+4 = 7

7+4 = 11
The square root of 11 not a whole number this is wrong😂😂😂

Well, proving this mathematically might not be the funny bone tickler you were hoping for, but I'll do my best to keep it entertaining!

Let's call the smaller consecutive positive integer "x" and the bigger one "x + 1". The product of these two numbers is x(x + 1).

Now, let's add the larger integer (x + 1) to the product: x(x + 1) + (x + 1).

If we simplify this expression, we get (x + 1)(x + 1), which can also be written as (x + 1)^2. And as you may know, a number raised to the power of 2 is just a fancy way of saying "square number"!

So, no matter what positive integers you choose, the result of multiplying the two consecutive positive integers and adding the larger one will always be a square number. Voila!

Hope that gave you a smile while proving the mathematical truth.

To prove that the result is always a square number, we need to consider two consecutive positive integers, let's call them n and n+1.

First, let's calculate the product of these two consecutive integers: n * (n+1).

Next, we will add the larger of the two integers, which is n+1, to the product: n * (n+1) + (n+1).

Now, we can simplify the expression: (n+1) * (n+1), which is equal to (n+1)².

The expression (n+1)² is the square of (n+1), so we have proven that the result is always a square number.

In mathematical notation, we can represent this as follows:
Let n be a positive integer.
(n * (n+1)) + (n+1) = (n+1)²

This proof holds true for any consecutive positive integers.