Show that $\sum _{j=1}^n{j}^2=\frac{n(n+1)(2n+1)}{6}.$

Step 1: Prove the base case (n=1) First, we need to plug n=1 into the formula: $\frac{1(1+1)(2(1)+1)}{6}=\frac{1(2)(3)}{6}=1$ Now we need to find the sum of the squares of the first 1 positive integers: $1^2=1$ Thus, the base case holds because the formula is true for n=1. Step 2: Define the induction step We need to assume that the statement is true for n=k, that is: $\sum _{j=1}^k{j}^2=\frac{k(k+1)(2k+1)}{6}$ Then, we will show that this also holds for n=k+1: $\sum _{j=1}^{k+1}{j}^2=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$ Step 3: Add the next term to the assumed statement (sum of squares up to k) We want to add the next term, $(k+1{)}^2$, to the sum of the squares up to k: $\sum _{j=1}^{k+1}{j}^2=\sum _{j=1}^k{j}^2+(k+1{)}^2$ We will substitute our assumption into this equation: $\frac{k(k+1)(2k+1)}{6}+(k+1{)}^2$ Step 4: Show that the sum of squares up to k+1 matches the formula for k+1 We need to show that the modified sum of squares matches the formula for k+1, which is $\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$. Let's simplify this expression and see if it matches: $\frac{k(k+1)(2k+1)}{6}+(k+1{)}^2=\frac{k(k+1)(2k+1)+6(k+1{)}^2}{6}$$L=\frac{(k+1)[k(2k+1)+6(k+1)]}{6}$$L=\frac{(k+1)(2k^2+k+6k+6)}{6}$$L=\frac{(k+1)(2k^2+7k+6)}{6}$ Now, let's simplify the formula for k+1: $R=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}=\frac{(k+1)(k+2)(2k+3)}{6}$ Comparing the two expressions, we see that $L=R$, which indicates that the statement holds for n=k+1. Step 5: Conclude the proof Since the base case (n=1) is true and the induction step has been shown to hold (if it's true for n=k, it's true for n=k+1), we can conclude that the sum of the squares of the first n positive integers is equal to $\frac{n(n+1)(2n+1)}{6}$ for all positive integers n.