prove algebraically that the difference between the squares of any two consecutive integers is equal to the sum of these two integers.
+28 1.
What the statement says:
\(({n+1}^{2})-{n}^{2} = 2n+1\)
Demonstration:
\(({n+1}^{2})-{n}^{2} = ({n}^{2}+2n+1)-{n}^{2}- ={n}^{2}+2n+1 -{n}^{2}- =2n+1\)
Example:
\({6}^{2}-{5}^{2}= 36-25=11\)
The sum of these two integers
\(6+5= 11\)
I hope it was helpful
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