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Suppose we have to prove that $1+3+5+...(2n-1)=n^2$

Here is my "proof," which I did by induction. Since I am learning induction, I wanted to know whether I did it correctly (I don't think I did).

For the base case, suppose n=1. Then $2(1)-1=1^2$, which is clearly true.

Now suppose $n=k$, then $1+3+5+...+ (2k-1)=k^2$

For $n=k+1$, $1+3+5+...+(2k+1)=(k+1)^2$. Then $1+3+5+...+(2k+1)=k^2+2k+1$. The 2k+1's cancel out, leaving $1+3+5+...+2k-1=k^2$, which is true by the induction hypothesis. QED?

UserM1
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    No, it is not correct. Your aim is to prove that $1+3+5+\cdots+(2k+1)=(k+1)^2$. What you did was to write “For $n=k=1$, $1+3+5+\cdots+(2k+1)=(k+1)^2$”, and then to obtain consequences of this equality. – José Carlos Santos Apr 16 '22 at 17:55
  • On the other hand I would say it is valid if the correct explanation is given along with the work. The 'Then' you have can be changed to a couple of 'If and only If' statements, which leads to your induction hypothesis. – David P Apr 16 '22 at 18:48
  • Additionally, this is not a duplicate, as this is tagged 'solution verification'. The OP has a specific question about their own solution. – David P Apr 16 '22 at 18:49

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