Solutions

(i)

$$\begin{aligned} (n + 1)^{4} - 1^{4} &= 4 (1^{3} + \dots + n^{3}) + 6 (1^{2} + \dots + n^{2}) + 4 (1 + \dots + n) + n \\ &= 4 \sum_{k = 0}^{n} k^{3} + 6 \frac{n (n + 1) (2n + 1)}{6} + 4 \frac{n (n + 1)}{2} + n\\ &= 4 \sum_{k = 0}^{n} k^{3} + n (n + 1) (2n + 1) + 2 n (n + 1) + n \end{aligned}$$

Moving the terms unrelated to the sum to the left, we obtain:

$$\begin{aligned} \sum_{k = 0}^{n} k^{3} &= \frac{{(n + 1)}^{4} - 1 - n (n + 1) (2n + 1) - 2n (n + 1) - n}{4} \\ &= \frac{n^{4} + 4n^{3} + 6n^{2} + 4n + 1 - 1 - 2n^{3} - 3n^{2} - n - 2n^{2} - 2n - n}{4} \\ &= \frac{n^{4} + 2n^{3} + n^{2}}{4} \\ &= \frac{n^{2} (n^{2} + 2n + 1)}{4} \\ &= \frac{n^{2} {(n + 1)}^{2} }{4} \\ \end{aligned}$$

(ii)

$$\begin{aligned} {(n + 1)}^{5} - 1^{5} &= 5 (1^{4} + \dots + n^{4}) + 10 (1^{3} + \dots + n^{3}) + 10 (1^{2} + \dots + n^{2}) \\ & \quad + 5(1 + \dots + n) + n \\ &= 5 \sum_{k = 0}^{n} k^{4} + 10 \frac{n^{2} {(n + 1)}^{2}}{4} + 10 \frac{n (n + 1) (2n + 1)}{6} + 5 \frac{n (n + 1)}{2} + n \end{aligned}$$

Again, we move all terms to the left except the sum.

$$\begin{aligned} \sum_{k = 0}^{n} k^{4} &= \frac{ {(n + 1)}^{5} - 1 - 10 \frac{n^{2} {(n + 1)}^{2}}{4} - 10 \frac{n (n + 1) (2n + 1)}{6} - 5 \frac{n (n + 1)}{2} - n} {5} \\ &= \frac{{(n + 1)}^{5}}{5} - \frac{1}{5} - 10 \frac{n^{2} {(n + 1)}^{2} }{20} - \frac{10 n (n + 1) (2n + 1)}{30} - 5 \frac{n (n + 1)}{10} - \frac{n}{5} \\ &= \frac{6 (n + 1)^{5} - 6 - 15 n^{2} (n + 1)^{2} - 10 n (n + 1) (2n + 1) - 15n (n + 1) - 6n}{30} \\ &= \frac{6n^{5} + 30n^{4} + 60n^{3} + 60n^{2} + 30n + 6 - 6 - 15n^{4} - 30n^{3} - 15n^{2} - 20 n^{3}}{30} \\ & \qquad \frac{- 30 n^{2} - 10n - 15 n^{2} - 15n - 6n}{30} \\ &= \frac{6n^{5} + 15n^{4} + 10n^{3} - n}{30} \\ &= \frac{n (6n^{4} + 15n^{3} + 10n^{2} - 1)}{30} \\ &= \frac{n (n + 1) (6n^{3} + 9n^{2} + n - 1)}{30} \\ &= \frac{n (n + 1) (2n + 1) (3n^{2} + 3n - 1)}{30} \end{aligned}$$

(iii) Notice that

$$\begin{aligned} \frac{1}{n (n + 1)} = \frac{1}{n} - \frac{1}{n + 1} \end{aligned}$$

Thus,

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

(iv) Notice that,

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

Thus,

$$\begin{aligned} \frac{3}{1^{2} \cdot 2^{2}} + \frac{5}{2^{2} \cdot 3^{2}} + \dots + \frac{2n + 1}{n^{2} (n + 1)^{2}} &= ( \frac{1}{1^{2}} - \frac{1}{2^{2}} ) + ( \frac{1}{2^{2}} - \frac{1}{3^{2}} ) + \dots + ( \frac{1}{n^{2}} - \frac{1}{(n + 1)^{2}} ) \\ &= \frac{1}{1} - \frac{1}{(n + 1)^{2}} \end{aligned}$$