已知 ∑r=1n7r+4r(r+1)(r+2)=92−2n+1−5n+2.\sum_{r=1}^{n} \dfrac{7r+4}{r(r+1)(r+2)} = \dfrac{9}{2} - \dfrac{2}{n+1} - \dfrac{5}{n+2}.

  1. (a)

    求 ∑r=72n7r−3r3−r\sum_{r=7}^{2n} \dfrac{7r-3}{r^3 - r}，用 nn 表示你的答案。

  2. (b)

    用代数方法证明，对于所有正整数 rr，都有 (r+1)3>r(r+1)(r+2)(r+1)^3 > r(r+1)(r+2)。

  3. (c)

    由此证明 ∑r=1n7r+4(r+1)3<92.\sum_{r=1}^{n} \dfrac{7r+4}{(r+1)^3} < \dfrac{9}{2}.

裂项相消求和与差分数列与级数概念分式不等式English
先不看答案，让 AI 诊断你卡在哪里？

它不会直接告诉你答案，而是问你卡在哪一步看看 AI 是怎么陪你一步步弄懂这道题的

完整解法

6(a)

1. Rewrite the general term of the summation by factorising its denominator.

7r−3r3−r=7r−3r(r2−1)=7r−3(r−1)r(r+1)\dfrac{7r-3}{r^3-r} = \dfrac{7r-3}{r(r^2-1)} = \dfrac{7r-3}{(r-1)r(r+1)}

2. Perform an index substitution to match the form of the given summation.

Let k=r−1. Then r=k+1.\text{Let } k = r-1. \text{ Then } r = k+1.
When r=7,k=6.\text{When } r = 7, k = 6.
When r=2n,k=2n−1.\text{When } r = 2n, k = 2n-1.
The term becomes 7(k+1)−3k(k+1)(k+2)=7k+4k(k+1)(k+2).\text{The term becomes } \dfrac{7(k+1)-3}{k(k+1)(k+2)} = \dfrac{7k+4}{k(k+1)(k+2)}.
Therefore, ∑r=72n7r−3r3−r=∑k=62n−17k+4k(k+1)(k+2).\text{Therefore, } \sum_{r=7}^{2n} \dfrac{7r-3}{r^3-r} = \sum_{k=6}^{2n-1} \dfrac{7k+4}{k(k+1)(k+2)}.

3. Express the sum starting from $k=6$ as the difference of two sums starting from $k=1$.

∑k=62n−17k+4k(k+1)(k+2)=∑k=12n−17k+4k(k+1)(k+2)−∑k=157k+4k(k+1)(k+2)\sum_{k=6}^{2n-1} \dfrac{7k+4}{k(k+1)(k+2)} = \sum_{k=1}^{2n-1} \dfrac{7k+4}{k(k+1)(k+2)} - \sum_{k=1}^{5} \dfrac{7k+4}{k(k+1)(k+2)}

4. Apply the given summation formula to both parts and simplify to obtain the final answer.

Using the given formula ∑r=1n7r+4r(r+1)(r+2)=92−2n+1−5n+2:\text{Using the given formula } \sum_{r=1}^{n} \dfrac{7r+4}{r(r+1)(r+2)} = \dfrac{9}{2} - \dfrac{2}{n+1} - \dfrac{5}{n+2}:
For n=2n−1:∑k=12n−1⋯=92−2(2n−1)+1−5(2n−1)+2=92−1n−52n+1\text{For } n = 2n-1: \quad \sum_{k=1}^{2n-1} \dots = \dfrac{9}{2} - \dfrac{2}{(2n-1)+1} - \dfrac{5}{(2n-1)+2} = \dfrac{9}{2} - \dfrac{1}{n} - \dfrac{5}{2n+1}
For n=5:∑k=15⋯=92−25+1−55+2=92−13−57\text{For } n = 5: \quad \sum_{k=1}^{5} \dots = \dfrac{9}{2} - \dfrac{2}{5+1} - \dfrac{5}{5+2} = \dfrac{9}{2} - \dfrac{1}{3} - \dfrac{5}{7}
Substituting back:\text{Substituting back:}
(92−1n−52n+1)−(92−13−57)=−1n−52n+1+13+57=2221−1n−52n+1\left( \dfrac{9}{2} - \dfrac{1}{n} - \dfrac{5}{2n+1} \right) - \left( \dfrac{9}{2} - \dfrac{1}{3} - \dfrac{5}{7} \right) = -\dfrac{1}{n} - \dfrac{5}{2n+1} + \dfrac{1}{3} + \dfrac{5}{7} = \dfrac{22}{21} - \dfrac{1}{n} - \dfrac{5}{2n+1}

答案

2221−1n−52n+1\dfrac{22}{21} - \dfrac{1}{n} - \dfrac{5}{2n+1}

6(b)

1. Find the difference between the left-hand side and the right-hand side of the inequality.

(r+1)3−r(r+1)(r+2)(r+1)^3 - r(r+1)(r+2)
=(r+1)[(r+1)2−r(r+2)]= (r+1)[(r+1)^2 - r(r+2)]
=(r+1)(r2+2r+1−r2−2r)= (r+1)(r^2 + 2r + 1 - r^2 - 2r)
=(r+1)(1)= (r+1)(1)
=r+1= r+1

2. Use the condition that $r$ is a positive integer to deduce the inequality.

Since rr is a positive integer (r≥1r \ge 1), we have r+1>0r+1 > 0.
Therefore,
(r+1)3−r(r+1)(r+2)>0(r+1)^3 - r(r+1)(r+2) > 0
(r+1)3>r(r+1)(r+2)(r+1)^3 > r(r+1)(r+2)

答案

(Shown)

6(c)

1. Deduce an inequality for the terms of the series using the result from (b).

From part (b), we have (r+1)3>r(r+1)(r+2)(r+1)^3 > r(r+1)(r+2).
Since r≥1r \ge 1, both sides are positive. Taking the reciprocal reverses the inequality:
1(r+1)3<1r(r+1)(r+2)\frac{1}{(r+1)^3} < \frac{1}{r(r+1)(r+2)}
Since 7r+4>07r+4 > 0 for r≥1r \ge 1, multiplying both sides gives:
7r+4(r+1)3<7r+4r(r+1)(r+2)\frac{7r+4}{(r+1)^3} < \frac{7r+4}{r(r+1)(r+2)}

2. Apply the summation to both sides and substitute the given series result.

Summing both sides from r=1r=1 to nn:
∑r=1n7r+4(r+1)3<∑r=1n7r+4r(r+1)(r+2)\sum_{r=1}^{n} \frac{7r+4}{(r+1)^3} < \sum_{r=1}^{n} \frac{7r+4}{r(r+1)(r+2)}
Substituting the expression given in the question stem:
∑r=1n7r+4(r+1)3<92−2n+1−5n+2\sum_{r=1}^{n} \frac{7r+4}{(r+1)^3} < \frac{9}{2} - \frac{2}{n+1} - \frac{5}{n+2}

3. Conclude the proof by evaluating the signs of the subtracted terms.

For all positive integers n≥1n \ge 1, the terms 2n+1\frac{2}{n+1} and 5n+2\frac{5}{n+2} are strictly positive.
Therefore, subtracting positive quantities from 92\frac{9}{2} results in a value strictly less than 92\frac{9}{2}:
92−2n+1−5n+2<92\frac{9}{2} - \frac{2}{n+1} - \frac{5}{n+2} < \frac{9}{2}
Hence, it follows that:
∑r=1n7r+4(r+1)3<92\sum_{r=1}^{n} \frac{7r+4}{(r+1)^3} < \frac{9}{2}

答案

Shown\text{Shown}

AI 分析，可能出错

用 AI 批改，圈出写错的那一步，并自动将你的薄弱点存入错题本，形成你的错因库。

同一知识点的其他题

  1. 1已知 f(r)=r(r−1)!f(r)=\dfrac{r}{(r-1)!}，证明 f(r−1)−f(r)=r2−3r+1(r−1)!f(r-1)-f(r)=\dfrac{r^2-3r+1}{(r-1)!}。 由此求 ∑r=2nr2−3r+1(r−1)!\displaystyle\sum_{r=2}^{n} \frac{r^2-3r+1}{(r-1)!}…
  2. 2已知数列{an}\{a_{n}\}中，a1=76a_{1}=\dfrac{7}{6}，an2−an+1=an+1(n∈N∗)a_{n}^{2}-a_{n}+1=a_{n+1}(n\in N^{*})，SnS_{n}是{1an}\{\dfrac{1}{a_{n}}\}的前…
  3. 3已知 f(r)=r(r−1)!f(r)=\dfrac{r}{(r-1)!}，证明 f(r−1)−f(r)=r2−3r+1(r−1)!f(r-1)-f(r)=\dfrac{r^2-3r+1}{(r-1)!}。 由此求 ∑r=2nr2−3r+1(r−1)!\displaystyle\sum_{r=2}^n \dfrac{r^2-3r+1}{(r-1)!}…
  4. 4证明 32r+1−42r+3+12r+5=8r+k(2r+1)(2r+3)(2r+5)\frac{3}{2r+1}-\frac{4}{2r+3}+\frac{1}{2r+5}=\frac{8r+k}{(2r+1)(2r+3)(2r+5)}，其中 kk 是待求常数。 由此求 ∑r=0n2r+7(2r+1)(2r+3)(2r+5)\sum_{r=0}^n \frac{2r+7}{(2r+1)(2r+3)(2r+5)}…
  5. 5用数学归纳法证明，对于所有正整数 nn 以及 x∈Rx\in\mathbb{R}，x≠0, x≠1x\ne 0,\ x\ne 1， 1x−1−1x−1x2−⋯−1xn=1xn(x−1) .\frac{1}{x-1}-\frac{1}{x}-\frac{1}{x^2}-\dots-\frac{1}{x^n}=\frac{1}{x^n(x-1)}\,.…