已知 ∑r=1n2r(r+1)(r+2)=12−1(n+1)(n+2)\displaystyle \sum_{r=1}^n \frac{2}{r(r+1)(r+2)} = \frac{1}{2} - \frac{1}{(n+1)(n+2)}。

  1. (a)
    1. (i)

      写出该级数的无穷和。

    2. (ii)

      求该级数前 nn 项的和

      217×18×19+218×19×20+219×20×21+⋯\displaystyle \frac{2}{17\times18\times19} + \frac{2}{18\times19\times20} + \frac{2}{19\times20\times21} + \cdots。

    3. (iii)

      求 ∑r=2n1r(r2−1)\displaystyle \sum_{r=2}^n \frac{1}{r(r^2-1)}。

  2. (b)

    数列 u1,u2,u3,…u_1,u_2,u_3,\dots 的定义为 u1=3u_1 = 3，un+1=2−4unu_{n+1} = 2 - \frac{4}{u_n}，n≥1n \ge 1。

    1. (i)

      求 u2u_2，u3u_3，u4u_4 和 u100u_{100} 的值。

    2. (ii)

      由此，用 nn 表示 ∑r=13n−2ur\displaystyle \sum_{r=1}^{3n-2} u_r，并化简你的答案。

级数收敛与无穷求和裂项相消求和与差分递推关系数列与级数概念通项公式English
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完整解法

9(a)(i)

1. Evaluate the limit of the sum to $n$ terms as $n$ approaches infinity.

lim⁡n→∞Sn=lim⁡n→∞(12−1(n+1)(n+2))\lim_{n \to \infty} S_n = \lim_{n \to \infty} \left( \frac{1}{2} - \frac{1}{(n+1)(n+2)} \right)

As n→∞n \to \infty, 1(n+1)(n+2)→0\dfrac{1}{(n+1)(n+2)} \to 0.

∴S∞=12\therefore S_\infty = \frac{1}{2}

答案

12\dfrac{1}{2}

9(a)(ii)

1. Express the required sum as the difference between two sums of the original series.

The series starts at r=17r=17. The nn-th term of this new series corresponds to r=16+nr=16+n.
Let the required sum be S′S'.
S′=∑r=17n+162r(r+1)(r+2)S' = \sum_{r=17}^{n+16} \frac{2}{r(r+1)(r+2)}
S′=∑r=1n+162r(r+1)(r+2)−∑r=1162r(r+1)(r+2)S' = \sum_{r=1}^{n+16} \frac{2}{r(r+1)(r+2)} - \sum_{r=1}^{16} \frac{2}{r(r+1)(r+2)}

2. Substitute the appropriate limits into the given sum formula and simplify.

Using the given formula ∑r=1n2r(r+1)(r+2)=12−1(n+1)(n+2)\sum_{r=1}^n \frac{2}{r(r+1)(r+2)} = \frac{1}{2} - \frac{1}{(n+1)(n+2)}:
S′=(12−1(n+16+1)(n+16+2))−(12−1(16+1)(16+2))S' = \left( \frac{1}{2} - \frac{1}{(n+16+1)(n+16+2)} \right) - \left( \frac{1}{2} - \frac{1}{(16+1)(16+2)} \right)
S′=(12−1(n+17)(n+18))−(12−117×18)S' = \left( \frac{1}{2} - \frac{1}{(n+17)(n+18)} \right) - \left( \frac{1}{2} - \frac{1}{17 \times 18} \right)
S′=117×18−1(n+17)(n+18)S' = \frac{1}{17 \times 18} - \frac{1}{(n+17)(n+18)}
S′=1306−1(n+17)(n+18)S' = \frac{1}{306} - \frac{1}{(n+17)(n+18)}

答案

1306−1(n+17)(n+18)\frac{1}{306} - \frac{1}{(n+17)(n+18)}

9(a)(iii)

1. Factorise the denominator of the general term to relate it to the given identity.

∑r=2n1r(r2−1)=∑r=2n1(r−1)r(r+1)\sum_{r=2}^n \frac{1}{r(r^2-1)} = \sum_{r=2}^n \frac{1}{(r-1)r(r+1)}

2. Adjust the index and introduce a constant factor to perfectly match the terms of the given series.

Let k=r−1k = r-1. As rr goes from 22 to nn, kk goes from 11 to n−1n-1.
∑r=2n1(r−1)r(r+1)=12∑r=2n2(r−1)r(r+1)\sum_{r=2}^n \frac{1}{(r-1)r(r+1)} = \frac{1}{2} \sum_{r=2}^n \frac{2}{(r-1)r(r+1)}
=12∑k=1n−12k(k+1)(k+2)= \frac{1}{2} \sum_{k=1}^{n-1} \frac{2}{k(k+1)(k+2)}

3. Apply the given sum formula by replacing $n$ with $n-1$.

Using the given identity ∑r=1m2r(r+1)(r+2)=12−1(m+1)(m+2)\displaystyle \sum_{r=1}^m \frac{2}{r(r+1)(r+2)} = \frac{1}{2} - \frac{1}{(m+1)(m+2)} with m=n−1m = n-1:
=12[12−1(n−1+1)(n−1+2)]= \frac{1}{2} \left[ \frac{1}{2} - \frac{1}{(n-1+1)(n-1+2)} \right]
=12[12−1n(n+1)]= \frac{1}{2} \left[ \frac{1}{2} - \frac{1}{n(n+1)} \right]
=14−12n(n+1)= \frac{1}{4} - \frac{1}{2n(n+1)}

答案

14−12n(n+1)\frac{1}{4} - \frac{1}{2n(n+1)}

9(b)(i)

1. Calculate the values of $u_2$, $u_3$, and $u_4$ using the recurrence relation

u2=2−4u1=2−43=23u_2 = 2 - \frac{4}{u_1} = 2 - \frac{4}{3} = \frac{2}{3}
u3=2−4u2=2−423=2−6=−4u_3 = 2 - \frac{4}{u_2} = 2 - \frac{4}{\frac{2}{3}} = 2 - 6 = -4
u4=2−4u3=2−4−4=2+1=3u_4 = 2 - \frac{4}{u_3} = 2 - \frac{4}{-4} = 2 + 1 = 3

2. Deduce the period of the sequence and evaluate $u_{100}$

Since u4=u1=3u_4 = u_1 = 3, the sequence is periodic with a repeating block of 3 terms: 3,23,−43, \frac{2}{3}, -4.
To find u100u_{100}, divide 100 by the period length 3:
100=33×3+1100 = 33 \times 3 + 1
The remainder is 1, so the 100th term is the 1st term in the repeating block.
u100=u1=3u_{100} = u_1 = 3

答案

u2=23u_2 = \frac{2}{3}, u3=−4u_3 = -4, u4=3u_4 = 3, u100=3u_{100} = 3

9(b)(ii)

1. Determine the first few terms (from part (b)(i)) to establish the periodicity of the sequence.

u1=3u_1 = 3
u2=2−43=23u_2 = 2 - \frac{4}{3} = \frac{2}{3}
u3=2−42/3=2−6=−4u_3 = 2 - \frac{4}{2/3} = 2 - 6 = -4
u4=2−4−4=3u_4 = 2 - \frac{4}{-4} = 3
The sequence repeats every 3 terms.

2. Calculate the sum of one complete cycle of 3 terms.

u1+u2+u3=3+23−4=−13u_1 + u_2 + u_3 = 3 + \frac{2}{3} - 4 = -\frac{1}{3}

3. Express the series sum using the number of full cycles and the remainder.

The number of terms is 3n−23n-2, which can be written as 3(n−1)+13(n-1) + 1.
This corresponds to (n−1)(n-1) full cycles of 3 terms, plus 11 additional term (which is u1u_1).
∑r=13n−2ur=(n−1)(u1+u2+u3)+u1\sum_{r=1}^{3n-2} u_r = (n-1)(u_1 + u_2 + u_3) + u_1
=(n−1)(−13)+3= (n-1)\left(-\frac{1}{3}\right) + 3
=−13n+13+3= -\frac{1}{3}n + \frac{1}{3} + 3
=10−n3= \frac{10 - n}{3}

答案

10−n3\frac{10 - n}{3}

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