已知一个数列满足 $u_{0}=2$ 且对于 $n\ge 1$，有 $u_{n}=u_{n-1}+n^{3}+\left(\dfrac{1}{2}\right)^{n}$。 (a) 已知 $\displaystyle \sum_{r=1}^{n} r^{3} = \dfrac{n^{2}(n+1)^{2}}{4}$。通过考虑 $\displaystyle \sum_{r=1}^{n}(u_{r}-u_{r-1})$，求出用 $n$ 表示的 $u_{n}$ 的表达式。 (b) 由此，利用在 (a) 中求得的 $u_{n}$ 表达式，求出 $\displaystyle \sum_{r=9}^{n} \left( (r+2)^{3} + \left(\dfrac{1}{2}\right)^{r+2} \right)$ 的精确值。

1. Rearrange the given recurrence relation to express $u_r - u_{r-1}$ and apply the summation from $1$ to $n$.

$u_{r} - u_{r-1} = r^{3} + \left(\frac{1}{2}\right)^{r}$

$\sum_{r=1}^{n}(u_{r}-u_{r-1}) = \sum_{r=1}^{n} \left( r^{3} + \left(\frac{1}{2}\right)^{r} \right)$

2. Evaluate the left side using the method of differences.

$\sum_{r=1}^{n}(u_{r}-u_{r-1}) = (u_1 - u_0) + (u_2 - u_1) + \dots + (u_n - u_{n-1})$
$= u_n - u_0$

3. Evaluate the right side by summing the $r^3$ term and the geometric progression.

$\sum_{r=1}^{n} r^{3} + \sum_{r=1}^{n} \left(\frac{1}{2}\right)^{r} = \frac{n^{2}(n+1)^{2}}{4} + \frac{\frac{1}{2}\left(1-\left(\frac{1}{2}\right)^{n}\right)}{1-\frac{1}{2}}$
$= \frac{n^{2}(n+1)^{2}}{4} + 1 - \left(\frac{1}{2}\right)^{n}$

4. Equate both sides, substitute the initial condition $u_0 = 2$, and isolate $u_n$.

$u_n - u_0 = \frac{n^{2}(n+1)^{2}}{4} + 1 - \left(\frac{1}{2}\right)^{n}$
$u_n - 2 = \frac{n^{2}(n+1)^{2}}{4} + 1 - \left(\frac{1}{2}\right)^{n}$
$u_n = \frac{n^{2}(n+1)^{2}}{4} + 3 - \left(\frac{1}{2}\right)^{n}$

1. Determine the explicit formula for $u_n$ from part (a) (since it is required to solve the target).

$u_n - u_0 = \sum_{r=1}^n (u_r - u_{r-1}) = \sum_{r=1}^n \left( r^3 + \left(\frac{1}{2}\right)^r \right)$
$u_n - 2 = \frac{n^2(n+1)^2}{4} + \frac{\frac{1}{2} \left( 1 - \left(\frac{1}{2}\right)^n \right)}{1 - \frac{1}{2}}$
$u_n = \frac{n^2(n+1)^2}{4} + 1 - \left(\frac{1}{2}\right)^n + 2 = \frac{n^2(n+1)^2}{4} + 3 - \left(\frac{1}{2}\right)^n$

2. Use an index substitution to express the required sum in terms of the terms of the sequence.

Let $k = r+2$. As $r$ varies from $9$ to $n$, $k$ varies from $11$ to $n+2$.
$\sum_{r=9}^n \left( (r+2)^3 + \left(\frac{1}{2}\right)^{r+2} \right) = \sum_{k=11}^{n+2} \left( k^3 + \left(\frac{1}{2}\right)^k \right)$
Since $u_k - u_{k-1} = k^3 + \left(\frac{1}{2}\right)^k$, the sum becomes:
$\sum_{k=11}^{n+2} (u_k - u_{k-1}) = u_{n+2} - u_{10}$

3. Substitute into the formula for $u_n$ and evaluate exactly.

$u_{n+2} - u_{10} = \left( \frac{(n+2)^2(n+3)^2}{4} + 3 - \left(\frac{1}{2}\right)^{n+2} \right) - \left( \frac{10^2(11)^2}{4} + 3 - \left(\frac{1}{2}\right)^{10} \right)$
$= \frac{(n+2)^2(n+3)^2}{4} - \left(\frac{1}{2}\right)^{n+2} - 3025 + \frac{1}{1024}$
$= \frac{(n+2)^2(n+3)^2}{4} - \left(\frac{1}{2}\right)^{n+2} - \frac{3097599}{1024}$