SHTQ cua day so

Question

Tim so hang tong quat cua day $(U_n)$:

$\left\{ \begin{array}{l} U_1=2\\ U_{n+1}=3U_n+n^2+1 \end{array} \right.$

score 5 · Answer 1

Viết lại công thức truy hồi $U_n=3U_{n-1}+(n-1)^2+1=3U_{n-1}+n^2-2n+2$

Mà $n^2-2n+2=\frac{-1}{2}n^2-\frac{1}{2}n-1+\frac{3}{2}(n-1)^2+\frac{3}{2}(n-1)+3$

$\Rightarrow U_n+\frac{1}{2}n^2+\frac{1}{2}n+1=3(U_{n-1}+\frac{1}{2}(n-1)^2+\frac{1}{2}(n-1)+1)$

$\Rightarrow V_n=3V_{n-1}=3^{n-1}.V_1=4.3^{n-1}$

$\Rightarrow U_n=4.3^{n-1}-\frac{n^2+n}{2}-1$

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