也想請教各位老師，關於17題除了估計，是否有嚴謹的做法說明是1.5

第 6 題
\(\begin{align}
  & {{a}^{2}}={{b}^{2}}+{{c}^{2}}-2bc\cos A \\
&  \\
& \overline{AD}=d \\
& {{b}^{2}}+{{c}^{2}}=2{{\left( \frac{1}{2}a \right)}^{2}}+2{{d}^{2}} \\
& {{d}^{2}}=\frac{{{b}^{2}}+{{c}^{2}}-\frac{1}{2}{{a}^{2}}}{2} \\
& =\frac{{{b}^{2}}+{{c}^{2}}-\frac{1}{2}\left( {{b}^{2}}+{{c}^{2}}-2bc\cos A \right)}{2} \\
& =\frac{{{b}^{2}}+{{c}^{2}}+2bc\cos A}{4} \\
&  \\
& {{\left( \Delta ABC \right)}^{2}}=4{{\left( \Delta ADC \right)}^{2}} \\
& \frac{1}{4}{{b}^{2}}{{c}^{2}}{{\sin }^{2}}A=4{{\left( \frac{1}{4}ad\sin \angle ADC \right)}^{2}}=\frac{1}{4}{{a}^{2}}{{d}^{2}}{{\sin }^{2}}\angle ADC \\
& {{\sin }^{2}}\angle ADC=\frac{{{b}^{2}}{{c}^{2}}{{\sin }^{2}}A}{{{a}^{2}}{{d}^{2}}} \\
& =\frac{{{b}^{2}}{{c}^{2}}{{\sin }^{2}}A}{\left( {{b}^{2}}+{{c}^{2}}-2bc\cos A \right)\left( \frac{{{b}^{2}}+{{c}^{2}}+2bc\cos A}{4} \right)} \\
& =\frac{4{{b}^{2}}{{c}^{2}}{{\sin }^{2}}A}{{{\left( {{b}^{2}}+{{c}^{2}} \right)}^{2}}-4{{b}^{2}}{{c}^{2}}{{\cos }^{2}}A} \\
& =\frac{4{{b}^{2}}{{c}^{2}}-4{{b}^{2}}{{c}^{2}}{{\cos }^{2}}A}{{{\left( {{b}^{2}}+{{c}^{2}} \right)}^{2}}-4{{b}^{2}}{{c}^{2}}{{\cos }^{2}}A}\quad \left( 0\le {{\cos }^{2}}A<1\ ,\ {{\left( {{b}^{2}}+{{c}^{2}} \right)}^{2}}\ge 4{{b}^{2}}{{c}^{2}} \right) \\
& \le \frac{4{{b}^{2}}{{c}^{2}}}{{{\left( {{b}^{2}}+{{c}^{2}} \right)}^{2}}} \\
&  \\
& \sin \angle ADC\le \frac{2bc}{{{b}^{2}}+{{c}^{2}}} \\
&  \\
\end{align}\)

[ 本帖最後由 thepiano 於 2023-4-11 18:13 編輯 ]

謝謝老師指點!

第15題 提供另一個作法

當年被105武陵慘虐之後，看到Math Pro的高手用這招，就一直熟記在心。

17. 令 \[S = \sum\limits_{k = 1}^n {\sqrt {k\left( {k + 2} \right)} }  = \sum\limits_{k = 1}^n {\sqrt {{{\left( {k + 1} \right)}^2} - 1} }  < \sum\limits_{k = 1}^n {\left( {k + 1} \right)}  = \frac{{n\left( {n + 3} \right)}}{2}\]
又由 GM \(\geq\) HM
\[S = \sum\limits_{k = 1}^n {\sqrt {k\left( {k + 2} \right)} }  \ge \sum\limits_{k = 1}^n {\frac{{k\left( {k + 2} \right)}}{{\left( {k + 1} \right)}}}  = \sum\limits_{k = 1}^n {\frac{{{{\left( {k + 1} \right)}^2} - 1}}{{\left( {k + 1} \right)}}}  = \sum\limits_{k = 1}^n {\left( {k + 1} \right)}  - \sum\limits_{k = 1}^n {\frac{1}{{k + 1}}} \]
因此  \[\frac{3}{2} - \frac{{\sum\limits_{k = 1}^n {\frac{1}{{k + 1}}} }}{n} < \frac{S}{n} - \frac{n}{2} < \frac{3}{2}\]

令 \[{H_n} = \sum\limits_{k = 1}^n {\frac{1}{{k + 1}}} \] ，由Stolz-Cesaro Theorem，
\[\mathop {\lim }\limits_{n \to \infty } \frac{{{H_{n + 1}} - {H_n}}}{{\left( {n + 1} \right) - n}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{{n + 2}} = 0 \Rightarrow \mathop {\lim }\limits_{n \to \infty } \frac{{{H_n}}}{n} = 0\]
         
所以\[\mathop {\lim }\limits_{n \to \infty } \frac{3}{2} - \frac{{\sum\limits_{k = 1}^n {\frac{1}{{k + 1}}} }}{n} = \mathop {\lim }\limits_{n \to \infty } \frac{3}{2} = \frac{3}{2}\] ，由夾擠定理，
\[\mathop {\lim }\limits_{n \to \infty } \frac{S}{n} - \frac{n}{2} = \frac{3}{2}\]

[ 本帖最後由 ouchbgb 於 2023-4-14 13:21 編輯 ]

引用:

原帖由 ouchbgb 於 2023-4-12 15:33 發表
17. 令 \[S = \sum\limits_{k = 1}^n {\sqrt {k\left( {k + 2} \right)} }  = \sum\limits_{k = 1}^n {\sqrt {{{\left( {k + 1} \right)}^2} - 1} }  < \sum\limits_{k = 1}^n {\left( {k + 1} \right)}  = \frac{{n ...

紅色處的不等式方向應是相反

謝謝指正, 抱歉.

想了很久，不知道這樣的寫法對不對

8題
代入幾項觀察可發現a1=1,a2=2,a4=3,a11=5....
數列1,2,4,7,11...猜測第k項為1+(1+2+...+k-1)=1+k(k-1)/2
令m=1+k(k-1)/2代入am可得am=k故得證

考試我是這樣寫，不確定有沒有給分


15題
設duece時最後假獲勝機率為P
則P=0.6*0.6+0.6*0.4P+0.4*0.6P (接下來兩局的所有可能情形)
可得P=9/13
所求為0.6P=27/65

[ 本帖最後由 no40508888 於 2023-4-13 11:50 編輯 ]

想請問第16題
我算出來x＝500*(4/3)^1000
不知是否正確