Cho dãy số [imath](u_n)[/imath] xác định bởi [imath]\begin{cases} u_1 = 2023 \\ u_{n+1} = \dfrac{u_n^4 + 2022^2}{u_n^3 - u_n + 4044} \end{cases}[/imath][imath]\forall n \in \mathbb{N}^{*}[/imath]
a)CMR: [imath]u_n > 2022, \forall n \in N^*[/imath]
b)Đặt [imath]S_n = \displaystyle\sum_{k=1}^{n} \dfrac{1}{u_k^3 + 2022},\forall n \in \mathbb{N}^*[/imath]
Tính [imath]lim S_n[/imath]
Thảo_UwU
b) [imath]u_{n+1} - 2022 = \dfrac{u_n^4 + 2022^2 - 2022u_n^3 + 2022u_n - 2.2022^2}{u_n^3 - u_n + 4044}[/imath]
[imath]\iff u_{n+1} - 2022 = \dfrac{u_n^4 + 2022^2 - 2022u_n^3 + 2022u_n - 2022^2 }{u_n^3 - u_n + 4044}[/imath]
[imath]\iff u_{n+1} - 2022 = \dfrac{ (u_n^3 + 2022)(u_n -2022)}{u_n^3 - u_n + 4044}[/imath]
[imath]\iff \dfrac{1}{u_{n+1} - 2022} = \dfrac{ u_n^3 +2022 - (u_n -2022) }{(u_n^3 + 2022)(u_n -2022)}[/imath]
[imath]\iff \dfrac{1}{u_{n+1} - 2022} = \dfrac{1}{u_{n} - 2022} - \dfrac{1}{u_n^3 + 2022}[/imath]
[imath]\iff \dfrac{1}{u_n^3 + 2022} = \dfrac{1}{u_{n} - 2022} - \dfrac{1}{u_{n +1} - 2022}[/imath]
Suy ra: [imath]S_n = \displaystyle\sum_{k=1}^{n} \dfrac{1}{u_k^3 + 2022} = \dfrac{1}{u_1 - 2022} - \dfrac{1}{u_2 - 2022} + \dfrac{1}{u_3 - 2022} - .... + \dfrac{1}{u_{n} - 2022} - \dfrac{1}{u_{n +1} - 2022} = \dfrac{1}{u_1 - 2022} - \dfrac{1}{u_{n +1} - 2022} [/imath]
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