Solutions

(i) We begin by proving that the base case holds; that the equality holds for $n = 1$.

$$\begin{aligned} 1^{2} &= \frac{1 (1 + 1)(2 \cdot 1 + 1)}{6} \\ &= \frac{1 \cdot 2 \cdot 3}{6} \\ &= \frac{6}{6} \\ &= 1 \end{aligned}$$

We then prove that the equation holds for $n+1$, if it holds for $n$.

$$\begin{aligned} 1^{2} + \dots + {(n + 1)}^{2} &= \frac{n(n+1)(2n+1)}{6} + {(n + 1)}^{2} \\ &= \frac{n(n+1)(2n+1)}{6} + \frac{6 {(n + 1)}^{2} }{6} \\ &= \frac{(n + 1)(n(2n + 1) + 6(n + 1))}{6} \\ &= \frac{(n + 1)(2n^{2} + n + 6n + 6)}{6} \\ &= \frac{(n + 1)(2n^{2} + 7n + 6)}{6} \\ &= \frac{(n + 1)(2n^{2} + 4n + 3n + 6)}{6} \\ &= \frac{(n + 1)(2n(n + 2) + 3 (n + 2))}{6} \\ &= \frac{(n + 1)((2n + 3)(n + 2))}{6} \\ &= \frac{(n + 1)(2n + 3)(n + 2)}{6} \end{aligned}$$

(ii) Again, we begin by proving the base case.

$$\begin{aligned} 1^{3} &= {(1)}^{2} \\ &= 1 \end{aligned}$$

We then move on to prove that the equation holds for $n + 1$, if it holds for $n$.

$$\begin{aligned} 1^{3} + \dots + {(n + 1)}^{3} &= {(1 + \dots + n)}^{2} + {(n + 1)}^{3} \\ &= ( \frac{n (n + 1)}{2} )^{2} + (n + 1)^{3} \\ &= \frac{n^{2} (n + 1)^{2} }{4} + \frac{4(n + 1)^{3}}{4} \\ &= \frac{(n + 1)^{2} (n^{2} + 4 (n + 1))}{4} \\ &= \frac{(n + 1)^{2} (n^{2} + 4n + 4)}{4} \\ &= \frac{(n + 1)^{2} (n + 2)^{2} }{4} \\ &= (\frac{(n + 1)(n + 2)}{2})^{2} \\ &= (1 + \dots + (n + 1) )^{2} \end{aligned}$$