Use mathematical induction to prove that $\sum _{j=1}^n(2j-1)=n^2$

Step 1: Base Case (n=1) Let's check the given expression for n=1: $\sum _{j=1}^1(2j-1)$ . The sum consists of only one term: $2(1)-1=1$ . Now, let's check if it equals to $1^2$ : $1^2=1$ . Since both expressions are equal, the base case is true. Step 2: Inductive Step - Assume true for n=k Assume that the statement holds for n = k, meaning that: $\sum _{j=1}^k(2j-1)=k^2$ Step 3: Show that it's true for n=k+1 Now we want to show that the statement also holds for n = k + 1. To do this, we add the (k+1)-th term to both sides of our assumption: $\sum _{j=1}^{k+1}(2j-1)=k^2+(2(k+1)-1)$ Let's simplify the expression on the right side: $k^2+(2k+2-1)=k^2+2k+1$ Now, we can use our assumption: $\sum _{j=1}^k(2j-1)+(2(k+1)-1)=k^2+2k+1$ Since our assumption says that $\sum _{j=1}^k(2j-1)=k^2$ , we have: $k^2+2k+1=(k+1{)}^2$ This is true, since $(k+1{)}^2$ can be expanded to $k^2+2k+1$ . Thus, the statement holds for n=k+1. Step 4: Conclusion Since the statement holds for the base case (n=1) and we proved that it holds for n=k+1 assuming it holds for n=k, by the principle of mathematical induction, the statement holds for all positive integers n: $\sum _{j=1}^n(2j-1)=n^2$