Prove by induction that $$1 + 3 + 5 + 7 + ... + (2n + 1) = (n+ 1)^2 $$ //for every n ∈ $\mathbb N$.
$$1+2+3+...+n=\frac{n(n+1)}2$$
Proof:
$$3+5+7+\ldots+(2n+1)=$$
$$=1+2+3+4+5+\ldots+(2n+1)+(2n+2)-1-2(1+2+3+\ldots+ n+n+1)=$$
$$=\frac{(2n+2)(2n+3)}2-1-2\frac{(n+1)(n+2)}2=(n+1)(2n+3)-1-(n+1)(n+2)=$$
$$=(n+1)\left[2n+3-n-2\right]-1=(n+1)^2-1=n^2+2n=n(n+2)$$
See that it is true for $n=1$ since $1(1+1)/2=1$ now supose that $1+2+3+...+n=\frac{n(n+1)}{2}$, let us prove for $n+1$, using the hypotesis of induction $$1+2+3+...+n+(n+1)=\frac{n(n+1)}{2}+n+1= (n+1)\left(\frac{n}{2}+1\right)=(n+1)\left(\frac{n+2}{2}\right)=\frac{(n+1)(n+2)}{2}$$ As desired.
That is not proving by induction at all. If you're proving by induction it would look something like this: Let S(n) be the statement that $\sum_{i=0}^n(2i+1)=(n+1)^2$ n=0, LHS=1, RHS=$1^2=1=LHS$
Assume that S(k) is true, i.e. $1+3+...+(2k+1)=(k+1)^2$ Must now prove S(k+1) is true, i.e. $1+3+5+...+(2k+1)+(2k+3)=(k+2)^2$
$LHS=1+3+5+...+(2k+1)+(2k+3)= (k+1)^2 +2k+3$, using the inductive hypothesis. $(k+1)^2+2k+3=k^2+2k+1+2k+3=k^2+4k+4=(k+2)^2=RHS$. So S(k+1) is true if S(k) is true. Since S(1) is true, true for all positive integers.
In proofs by induction, we first test the base case (the initial starting case). We then assume it holds true for some value- the inductive hypothesis. Using this inductive hypothesis, we then show that the next value is true. As the base case is true, then it must be true for the subsequent values as well.
