Answer:
Step-by-step explanation:
It is true that for any given odd integer, square of that integer will also be odd.
i.e if [tex]m[/tex] is and odd integer then [tex]m^{2}[/tex] is also odd.
In the given proof the expansion for [tex](2k + 1)^{2}[/tex] is incorrect.
By definition we know,
[tex](a+b)^{2} = a^{2} + b^{2} + 2ab[/tex]
∴ [tex](2k + 1)^{2} = (2k)^{2} + 1^{2} + 2(2k)(1)\\(2k + 1)^{2} = 4k^{2} + 1 + 4k[/tex]
Now, we know [tex]4k^{2}[/tex] and [tex]4k[/tex] will be even values
∴[tex]4k^{2} + 1 + 4k[/tex] will be odd
hence [tex](2k + 1)^{2}[/tex] will be odd, which means [tex]m^{2}[/tex] will be odd.