已知 f(r)=1−3(r+2)!\mathrm{f}(r) = 1 - \frac{3}{(r+2)!}，其中 rr 是非负整数。 证明 f(r)−f(r−1)=a(r+1)(r+2)!\mathrm{f}(r) - \mathrm{f}(r-1) = \frac{a(r+1)}{(r+2)!}，其中 aa 是待确定的常数。

  1. (i)

    利用上述结果，求出用 nn 表示的 ∑r=1nr+1(r+2)!\sum_{r=1}^{n} \frac{r+1}{(r+2)!} 的表达式。

  2. (ii)

    利用第 (i) 部分的结果，推导 ∑r=3nr−1r!\sum_{r=3}^{n} \frac{r-1}{r!} 的值，并将你的结果化简。

裂项相消求和与差分通项公式数列与函数的关系English
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完整解法

4(i)

1. Determine the constant $a$ from the preliminary part and express the summand as a difference.

Since the previous part asks to determine aa, we first evaluate f(r)−f(r−1)\mathrm{f}(r) - \mathrm{f}(r-1):
f(r)−f(r−1)=(1−3(r+2)!)−(1−3(r+1)!)\mathrm{f}(r) - \mathrm{f}(r-1) = \left( 1 - \frac{3}{(r+2)!} \right) - \left( 1 - \frac{3}{(r+1)!} \right)
=3(r+1)!−3(r+2)!=3(r+2)−3(r+2)!=3(r+1)(r+2)!= \frac{3}{(r+1)!} - \frac{3}{(r+2)!} = \frac{3(r+2) - 3}{(r+2)!} = \frac{3(r+1)}{(r+2)!}
Thus, a=3a = 3, which gives r+1(r+2)!=13[f(r)−f(r−1)]\frac{r+1}{(r+2)!} = \frac{1}{3}[\mathrm{f}(r) - \mathrm{f}(r-1)].

2. Set up the summation and apply the method of differences.

∑r=1nr+1(r+2)!=13∑r=1n[f(r)−f(r−1)]\sum_{r=1}^{n} \frac{r+1}{(r+2)!} = \frac{1}{3} \sum_{r=1}^{n} [\mathrm{f}(r) - \mathrm{f}(r-1)]
=13[(f(1)−f(0))+(f(2)−f(1))+⋯+(f(n)−f(n−1))]= \frac{1}{3} \Big[ (\mathrm{f}(1) - \mathrm{f}(0)) + (\mathrm{f}(2) - \mathrm{f}(1)) + \dots + (\mathrm{f}(n) - \mathrm{f}(n-1)) \Big]

3. Cancel intermediate terms and substitute the function values.

Cancelling the intermediate terms leaves only the first and last terms:
=13[f(n)−f(0)]= \frac{1}{3} [\mathrm{f}(n) - \mathrm{f}(0)]
Substituting f(n)=1−3(n+2)!\mathrm{f}(n) = 1 - \frac{3}{(n+2)!} and f(0)=1−32!=−12\mathrm{f}(0) = 1 - \frac{3}{2!} = -\frac{1}{2}:
=13[(1−3(n+2)!)−(−12)]= \frac{1}{3} \left[ \left( 1 - \frac{3}{(n+2)!} \right) - \left(-\frac{1}{2}\right) \right]

4. Simplify the final algebraic expression.

=13[32−3(n+2)!]= \frac{1}{3} \left[ \frac{3}{2} - \frac{3}{(n+2)!} \right]
=12−1(n+2)!= \frac{1}{2} - \frac{1}{(n+2)!}

答案

12−1(n+2)!\frac{1}{2} - \frac{1}{(n+2)!}

4(ii)

1. Concisely calculate the result of part (i) as it is required to deduce the final answer.

From the main question stem, we first find aa:
f(r)−f(r−1)=(1−3(r+2)!)−(1−3(r+1)!)\mathrm{f}(r) - \mathrm{f}(r-1) = \left( 1 - \frac{3}{(r+2)!} \right) - \left( 1 - \frac{3}{(r+1)!} \right)
=3(r+1)!−3(r+2)!=3(r+2)−3(r+2)!=3(r+1)(r+2)!= \frac{3}{(r+1)!} - \frac{3}{(r+2)!} = \frac{3(r+2) - 3}{(r+2)!} = \frac{3(r+1)}{(r+2)!}
Thus, a=3a = 3. For part (i), using the method of differences:
∑r=1nr+1(r+2)!=13∑r=1n[f(r)−f(r−1)]\sum_{r=1}^{n} \frac{r+1}{(r+2)!} = \frac{1}{3} \sum_{r=1}^{n} [\mathrm{f}(r) - \mathrm{f}(r-1)]
=13[f(n)−f(0)]= \frac{1}{3} [\mathrm{f}(n) - \mathrm{f}(0)]
=13[(1−3(n+2)!)−(1−32)]=12−1(n+2)!= \frac{1}{3} \left[ \left(1 - \frac{3}{(n+2)!}\right) - \left(1 - \frac{3}{2}\right) \right] = \frac{1}{2} - \frac{1}{(n+2)!}

2. Use an index substitution to relate the sum in part (ii) to the sum in part (i).

We need to evaluate ∑r=3nr−1r!\sum_{r=3}^{n} \frac{r-1}{r!}.
Let k=r−2k = r - 2. The limits change accordingly:
When r=3r = 3, k=1k = 1.
When r=nr = n, k=n−2k = n - 2.
Substituting this into the summation:
∑r=3nr−1r!=∑k=1n−2(k+2)−1(k+2)!\sum_{r=3}^{n} \frac{r-1}{r!} = \sum_{k=1}^{n-2} \frac{(k+2)-1}{(k+2)!}
=∑k=1n−2k+1(k+2)!= \sum_{k=1}^{n-2} \frac{k+1}{(k+2)!}

3. Apply the result from part (i) to deduce the final value.

Using the established result ∑r=1Nr+1(r+2)!=12−1(N+2)!\sum_{r=1}^{N} \frac{r+1}{(r+2)!} = \frac{1}{2} - \frac{1}{(N+2)!} with the new upper limit N=n−2N = n - 2:
∑k=1n−2k+1(k+2)!=12−1((n−2)+2)!\sum_{k=1}^{n-2} \frac{k+1}{(k+2)!} = \frac{1}{2} - \frac{1}{((n-2)+2)!}
=12−1n!= \frac{1}{2} - \frac{1}{n!}

答案

12−1n!\frac{1}{2} - \frac{1}{n!}

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