➷ The first square number can be represented by [tex]n^{2}[/tex]
The second square number can be represented by [tex](n+1)^{2}[/tex]
Subtract them:
[tex](n+1)^{2} - n^{2}[/tex]
Expand it first:
[tex]n^{2} + 2n + 1 - n^{2}[/tex]
Simplify:
[tex]2n+1[/tex]
Anything multiplied by 2 is even. Adding 1 to this value would make it odd. This proves it.
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Answer:
[tex](a+1)^{2} -a^{2} = \\a^{2}+1+2a-a^{2} =\\ 2a+1[/tex]
It's always odd
Step-by-step explanation:
