Problem Detail: $1+3+5+…+(2n+3)=n^2+4n$ 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.
Asked By : user5507
Answered By : yuanshang
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 $n$.
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Question Source : http://cs.stackexchange.com/questions/21593
