Prove algebraically that the difference between the square of any two consecutive integers is equal to the sum of these two integers.

First number = n
Second number = n+1

Square the second number: (?????)^2

Difference between the squares: ?????

Sum of the consecutive integers = ? + ? = ?

Thank you.

If the two numbers are n and n+1, then we have

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

To prove algebraically that the difference between the square of any two consecutive integers is equal to the sum of these two integers, let's follow these steps:

Step 1: Define the two consecutive integers as n and n+1.

Step 2: Calculate the square of the second number, which is (n+1)^2.

By expanding the expression (n+1)^2, we get:
(n+1)^2 = n^2 + 2n + 1

Step 3: Calculate the square of the first number, which is n^2.

Step 4: Find the difference between the squares of the two numbers.

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

Step 5: Calculate the sum of the two consecutive integers.

Sum = n + (n+1)
Sum = 2n + 1

Step 6: Compare the difference between the squares of the two numbers to their sum.

Difference = 2n + 1
Sum = 2n + 1

Since the expressions for the difference and the sum are the same, we can conclude that the difference between the square of any two consecutive integers is equal to the sum of these two integers.

Therefore, algebraically, the statement is proven.

To prove algebraically that the difference between the square of any two consecutive integers is equal to the sum of these two integers, let's follow the steps below:

Step 1: Assign variables to the consecutive integers.
Let the first number be represented by 'n', and the second number be 'n+1', since the second integer is always one greater than the first.

Step 2: Square the second number.
To square the second number (n+1), you need to multiply it by itself: (n+1)^2.

Step 3: Find the square of the second number.
Using the identity (a+b)^2 = a^2 + 2ab + b^2, we can expand (n+1)^2 as follows: (n+1)(n+1) = n^2 + 2n + 1.

Step 4: Find the square of the first number.
To square the first number (n), you need to multiply it by itself: n^2.

Step 5: Find the difference between the squares.
To find the difference between the squares, subtract the square of the first number from the square of the second number: (n^2 + 2n + 1) - n^2 = 2n + 1.

Step 6: Find the sum of the consecutive integers.
The sum of the two consecutive integers is given by n + (n+1) = 2n + 1.

Step 7: Compare the difference between the squares and the sum of the consecutive integers.
By comparing the difference obtained in Step 5, which is 2n + 1, and the sum obtained in Step 6, which is also 2n + 1, we can see that they are equal.

Therefore, algebraically, we have proven that the difference between the square of any two consecutive integers is equal to the sum of these two integers.