1. **Statement of the problem:** We have two sequences $(u_n)$ and $(v_n)$ defined on $\mathbb{N}$ with given conditions and formulas. We need to solve several questions about these sequences. --- ### Exercise 1 2. Given system: $$\begin{cases} u_0 + u_1 = 10 \\ u_2 + u_3 + u_4 = 27 \end{cases}$$ 3. $(u_n)$ is an arithmetic sequence, so: $$u_n = u_0 + n r$$ where $r$ is the common difference (the "basis"), and $u_0$ is the first term. 4. Express terms: $$u_1 = u_0 + r$$ $$u_2 = u_0 + 2r$$ $$u_3 = u_0 + 3r$$ $$u_4 = u_0 + 4r$$ 5. Use the system: - From $u_0 + u_1 = 10$: $$u_0 + (u_0 + r) = 10 \implies 2u_0 + r = 10$$ - From $u_2 + u_3 + u_4 = 27$: $$ (u_0 + 2r) + (u_0 + 3r) + (u_0 + 4r) = 27 \implies 3u_0 + 9r = 27$$ 6. Solve the system: From first: $$r = 10 - 2u_0$$ Substitute into second: $$3u_0 + 9(10 - 2u_0) = 27$$ $$3u_0 + 90 - 18u_0 = 27$$ $$-15u_0 = 27 - 90 = -63$$ $$u_0 = \frac{63}{15} = 4.2$$ Then: $$r = 10 - 2 \times 4.2 = 10 - 8.4 = 1.6$$ 7. Calculate $u_1$ and $u_3$: $$u_1 = u_0 + r = 4.2 + 1.6 = 5.8$$ $$u_3 = u_0 + 3r = 4.2 + 3 \times 1.6 = 4.2 + 4.8 = 9$$ 8. Check formula $u_n = 3 + 2n$: Given $u_0 = 4.2$ and $r=1.6$, this does not match $3 + 2n$. So the formula is not correct for this sequence. 9. Check if 4035 is a term of $(u_n)$: $$u_n = u_0 + n r = 4.2 + 1.6 n$$ Set equal to 4035: $$4.2 + 1.6 n = 4035 \implies 1.6 n = 4030.8 \implies n = \frac{4030.8}{1.6} = 2519.25$$ Since $n$ is not an integer, 4035 is not a term of the sequence. 10. Calculate sum $S = u_{35} + \cdots + u_7 + u_6$: Sum from $u_6$ to $u_{35}$ has $35 - 6 + 1 = 30$ terms. Sum of arithmetic sequence: $$S = \frac{number\ of\ terms}{2} \times (first\ term + last\ term)$$ Calculate $u_6$ and $u_{35}$: $$u_6 = 4.2 + 1.6 \times 6 = 4.2 + 9.6 = 13.8$$ $$u_{35} = 4.2 + 1.6 \times 35 = 4.2 + 56 = 60.2$$ Sum: $$S = \frac{30}{2} \times (13.8 + 60.2) = 15 \times 74 = 1110$$ --- ### Exercise 2 Given: $$u_n = 2 - 6n$$ $$v_n = 4 \times 3^n$$ 1. Calculate $u_0, u_1, v_0, v_1$: $$u_0 = 2 - 6 \times 0 = 2$$ $$u_1 = 2 - 6 \times 1 = 2 - 6 = -4$$ $$v_0 = 4 \times 3^0 = 4 \times 1 = 4$$ $$v_1 = 4 \times 3^1 = 4 \times 3 = 12$$ 2. (a) Show $(u_n)$ is arithmetic: Difference: $$u_{n+1} - u_n = (2 - 6(n+1)) - (2 - 6n) = 2 - 6n - 6 - 2 + 6n = -6$$ Constant difference $r = -6$, so $(u_n)$ is arithmetic with basis $-6$. (b) Check if $-11770$ is a term of $(u_n)$: $$u_n = 2 - 6n = -11770$$ $$-6n = -11772 \implies n = 1962$$ Since $n$ is a natural number, $-11770$ is the term $u_{1962}$. (c) Sum $S_n = u_0 + u_1 + \cdots + u_n$: Sum of arithmetic sequence: $$S_n = (n+1) \times \frac{u_0 + u_n}{2}$$ Calculate $u_n$: $$u_n = 2 - 6n$$ So: $$S_n = (n+1) \times \frac{2 + (2 - 6n)}{2} = (n+1) \times \frac{4 - 6n}{2} = (n+1)(2 - 3n)$$ 3. (a) Show $(v_n)$ is geometric: Ratio: $$\frac{v_{n+1}}{v_n} = \frac{4 \times 3^{n+1}}{4 \times 3^n} = 3$$ So $(v_n)$ is geometric with ratio $q=3$. (b) Show: $$v_{n+1} - v_n = 8 \times 3^n$$ Calculate: $$v_{n+1} - v_n = 4 \times 3^{n+1} - 4 \times 3^n = 4 \times 3^n (3 - 1) = 4 \times 3^n \times 2 = 8 \times 3^n$$ True. (c) Sum $S'_n = v_0 + v_1 + \cdots + v_n$: Sum of geometric sequence: $$S'_n = v_0 \times \frac{q^{n+1} - 1}{q - 1} = 4 \times \frac{3^{n+1} - 1}{3 - 1} = 4 \times \frac{3^{n+1} - 1}{2} = 2(3^{n+1} - 1)$$ --- **Final answers:** - Exercise 1: - $u_1 = 5.8$, $u_3 = 9$ - $u_0 = 4.2$, $r = 1.6$ - Formula $u_n = 3 + 2n$ is incorrect - 4035 is not a term - Sum $S = 1110$ - Exercise 2: - $u_0=2$, $u_1=-4$, $v_0=4$, $v_1=12$ - $(u_n)$ arithmetic with $r=-6$ - $-11770$ is term $u_{1962}$ - $S_n = (n+1)(2 - 3n)$ - $(v_n)$ geometric with ratio $3$ - $v_{n+1} - v_n = 8 \times 3^n$ - $S'_n = 2(3^{n+1} - 1)$