已知一个数列满足 $u_0 = 2$ ，且对于 $n \ge 1$ 有 $u_n = u_{n-1} + n^3 + \left(\frac{1}{2}\right)^n$ 。

(a)
已知 $\sum_{r=1}^{n} r^3 = \frac{n^2 (n+1)^2}{4}$ 。通过考虑 $\sum_{r=1}^{n} (u_r - u_{r-1})$ ，求用 $n$ 表示 $u_n$ 的公式。
(b)
因此，利用在 (a) 中求得的 $u_n$ 公式，求 $\sum_{r=9}^{n} \left[ (r+2)^3 + \left(\frac{1}{2}\right)^{r+2} \right]$ 的精确值。

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完整解法

4(a)

1. Set up the summation equation from the given recurrence relation.

Rearranging the given relation:
$u_r - u_{r-1} = r^3 + \left(\frac{1}{2}\right)^r$

Summing both sides from $r=1$ to $n$ :
$\sum_{r=1}^{n} (u_r - u_{r-1}) = \sum_{r=1}^{n} \left[ r^3 + \left(\frac{1}{2}\right)^r \right]$

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

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

Since $u_0 = 2$ ,
$\sum_{r=1}^{n} (u_r - u_{r-1}) = u_n - 2$

3. Evaluate the right-hand side using the given formula and the sum of a geometric series.

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

4. Equate both sides to find the explicit formula for $u_n$.

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

答案

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

4(b)

1. Determine the expression for $u_n$ from part (a) (needed for the subsequent steps).

\begin{aligned} \sum_{r=1}^{n} (u_r - u_{r-1}) &= u_n - u_0 = u_n - 2 \\ \text{Also, } \sum_{r=1}^{n} (u_r - u_{r-1}) &= \sum_{r=1}^{n} \left[ r^3 + \left(\frac{1}{2}\right)^r \right] \\ &= \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 \\ \therefore u_n &= \frac{n^2(n+1)^2}{4} + 3 - \left(\frac{1}{2}\right)^n \end{aligned}

2. Express the given summation's terms using the recurrence relation.

\begin{aligned} \text{From the given sequence, } u_r - u_{r-1} &= r^3 + \left(\frac{1}{2}\right)^r \\ \text{Substituting } r \text{ with } r+2\text{:} \\ u_{r+2} - u_{r+1} &= (r+2)^3 + \left(\frac{1}{2}\right)^{r+2} \end{aligned}

3. Apply the method of differences to simplify the summation.

\begin{aligned} \sum_{r=9}^{n} \left[ (r+2)^3 + \left(\frac{1}{2}\right)^{r+2} \right] &= \sum_{r=9}^{n} (u_{r+2} - u_{r+1}) \\ &= (u_{11} - u_{10}) + (u_{12} - u_{11}) + \dots + (u_{n+2} - u_{n+1}) \\ &= u_{n+2} - u_{10} \end{aligned}

4. Substitute the formula for $u_n$ to obtain the exact value.

\begin{aligned} 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{1}{4}(n+2)^2(n+3)^2 - \left(\frac{1}{2}\right)^{n+2} - 3025 + \frac{1}{1024} \\ &= \frac{1}{4}(n+2)^2(n+3)^2 - \left(\frac{1}{2}\right)^{n+2} - \frac{3097599}{1024} \end{aligned}

答案

$\frac{1}{4}(n+2)^2(n+3)^2 - \left(\frac{1}{2}\right)^{n+2} - \frac{3097599}{1024}$

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已知一个数列满足 u0=2u_0 = 2u0​=2，且对于 n≥1n \ge 1n≥1 有 un=un−1+n3+(12)nu_n = u_{n-1} + n^3 + \left(\frac{1}{2}\right)^nun​=un−1​+n3+(21​)n。

完整解法

4(a)

4(b)

同一知识点的其他题

已知一个数列满足 $u_0 = 2$ ，且对于 $n \ge 1$ 有 $u_n = u_{n-1} + n^3 + \left(\frac{1}{2}\right)^n$ 。