$1+3+5+...+(2n+3)=n^2+4n$For this series using induction proof. Base case 1,2,3,.. not working. But induction step works well. Base case is not given in question.
Induction proof, base case not working but induction step works?
proof techniques;induction
First of all, the right hand side should be $n^2+4n+4$You can find the generalized term for the left hand side is $2n-1$. Therefore, $1+3+5+...+(2n-1)+(2n+1)+(2n+3)=n^2+4n+4$, which gives you:$1+3+5+...+(2n-1)=n^2$. Then you can apply the mathematical induction to prove it holds for every positive integer $n$.