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