1. **Problem statement:** Calculate the first three terms of each sequence $(u_n)_n$ defined by the given general term, then find expressions for $u_{2n}$ and $u_{n+1}$. --- 2. **Sequence 1:** $u_n = n^2 + 2n + 3$ - First three terms: - $u_1 = 1^2 + 2\times1 + 3 = 1 + 2 + 3 = 6$ - $u_2 = 2^2 + 2\times2 + 3 = 4 + 4 + 3 = 11$ - $u_3 = 3^2 + 2\times3 + 3 = 9 + 6 + 3 = 18$ - Expressions: - $u_{2n} = (2n)^2 + 2(2n) + 3 = 4n^2 + 4n + 3$ - $u_{n+1} = (n+1)^2 + 2(n+1) + 3 = n^2 + 2n + 1 + 2n + 2 + 3 = n^2 + 4n + 6$ --- 3. **Sequence 2:** $u_n = \frac{2n - (-1)^{n+1}}{5n + 2}$ - First three terms: - $u_1 = \frac{2\times1 - (-1)^2}{5\times1 + 2} = \frac{2 - 1}{7} = \frac{1}{7}$ - $u_2 = \frac{4 - (-1)^3}{10 + 2} = \frac{4 + 1}{12} = \frac{5}{12}$ - $u_3 = \frac{6 - (-1)^4}{15 + 2} = \frac{6 - 1}{17} = \frac{5}{17}$ - Expressions: - $u_{2n} = \frac{4n - (-1)^{2n+1}}{10n + 2}$ - $u_{n+1} = \frac{2(n+1) - (-1)^{n+2}}{5(n+1) + 2} = \frac{2n + 2 - (-1)^{n+2}}{5n + 7}$ --- 4. **Sequence 3:** $u_n = \sum_{k=0}^n (-1)^k$ - First three terms: - $u_0 = (-1)^0 = 1$ - $u_1 = 1 + (-1)^1 = 1 - 1 = 0$ - $u_2 = 0 + (-1)^2 = 0 + 1 = 1$ - Expressions: - $u_{2n} = \sum_{k=0}^{2n} (-1)^k$ - $u_{n+1} = \sum_{k=0}^{n+1} (-1)^k$ --- 5. **Sequence 4:** $u_n = 13 + 2^3 + 3^3 + \cdots + n^3$ - Note: $2^3 = 8$, so the sum is $13 + \sum_{k=2}^n k^3$ - First three terms: - $u_1 = 13$ - $u_2 = 13 + 8 = 21$ - $u_3 = 21 + 27 = 48$ - Expressions: - $u_{2n} = 13 + \sum_{k=2}^{2n} k^3$ - $u_{n+1} = 13 + \sum_{k=2}^{n+1} k^3$ --- 6. **Sequence 5:** $u_n = 2 \times 4 \times 6 \times \cdots \times (2n)$ (product of even numbers up to $2n$) - First three terms: - $u_1 = 2$ - $u_2 = 2 \times 4 = 8$ - $u_3 = 2 \times 4 \times 6 = 48$ - Expressions: - $u_{2n} = \prod_{k=1}^{2n} 2k$ - $u_{n+1} = \prod_{k=1}^{n+1} 2k$ --- 7. **Sequence 6:** $u_n = 1 \times 3 \times 5 \times \cdots \times (2n - 1)$ (product of odd numbers up to $2n-1$) - First three terms: - $u_1 = 1$ - $u_2 = 1 \times 3 = 3$ - $u_3 = 1 \times 3 \times 5 = 15$ - Expressions: - $u_{2n} = \prod_{k=1}^{2n} (2k - 1)$ - $u_{n+1} = \prod_{k=1}^{n+1} (2k - 1)$ --- 8. **Sequence 7:** $u_n = \frac{1}{n^2 + 1} + \frac{1}{n^2 + 2} + \cdots + \frac{1}{n^2 + n}$ - First three terms: - $u_1 = \frac{1}{1 + 1} = \frac{1}{2}$ - $u_2 = \frac{1}{4 + 1} + \frac{1}{4 + 2} = \frac{1}{5} + \frac{1}{6} = \frac{11}{30}$ - $u_3 = \frac{1}{9 + 1} + \frac{1}{9 + 2} + \frac{1}{9 + 3} = \frac{1}{10} + \frac{1}{11} + \frac{1}{12} = \frac{131}{1320}$ - Expressions: - $u_{2n} = \sum_{k=1}^{2n} \frac{1}{(2n)^2 + k}$ - $u_{n+1} = \sum_{k=1}^{n+1} \frac{1}{(n+1)^2 + k}$ --- **Summary:** - Calculated first three terms for each sequence. - Expressed $u_{2n}$ and $u_{n+1}$ by substituting $n$ with $2n$ and $n+1$ respectively. Final answers are as above for each sequence.