Questions: Prove: The difference of the squares of two successive integers is equal to the sum of the integers. (n+1)^2-n^2 = [?] n+[ ] = n+(n+1)

Prove: The difference of the squares of two successive integers is equal to the sum of the integers.
(n+1)^2-n^2  = [?] n+[ ] 
 = n+(n+1)
Transcript text: Visible 9:30 AM Number Theory Prove: The difference of the squares of two successive integers is equal to the sum of the integers. \[ \begin{aligned} (n+1)^{2}-n^{2} & =[?] n+[\square \\ & =n+(n+1) \end{aligned} \]
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Solution

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Solution Steps

To prove that the difference of the squares of two successive integers is equal to the sum of the integers, we can start by expressing the two successive integers as \( n \) and \( n+1 \). Then, calculate the square of each and find their difference. Simplify the expression to show that it equals the sum of the integers \( n \) and \( n+1 \).

Step 1: Define the Integers

Let \( n \) be an integer. The two successive integers can be expressed as \( n \) and \( n + 1 \).

Step 2: Calculate the Difference of Squares

The difference of the squares of these integers is given by: \[ (n + 1)^2 - n^2 \] Calculating this, we have: \[ (n + 1)^2 = n^2 + 2n + 1 \] Thus, \[ (n + 1)^2 - n^2 = (n^2 + 2n + 1) - n^2 = 2n + 1 \]

Step 3: Calculate the Sum of the Integers

The sum of the integers \( n \) and \( n + 1 \) is: \[ n + (n + 1) = 2n + 1 \]

Step 4: Compare the Results

From Steps 2 and 3, we see that: \[ (n + 1)^2 - n^2 = 2n + 1 = n + (n + 1) \] This confirms that the difference of the squares of two successive integers is indeed equal to the sum of the integers.

Final Answer

The difference of the squares of two successive integers is equal to the sum of the integers, which can be expressed as: \[ \boxed{2n + 1} \]

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