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&lt;p&gt;前四項為
$$a_{1}=10,a_{2}=10^{2},a_{3}=10^{3},a_{4}=10^{4}$$
，接著把sigma展開得&lt;/p&gt;
&lt;p&gt;$$b=\sum_{n=1}^{3}log_{a_{n}}a_{n+1}=log_{a_{1}}a_{2}+log_{a_{2}}a_{3}+log_{a_{3}}a_{4}$$&lt;/p&gt;
&lt;p&gt;$$\Rightarrow b=log_{10}10^{2}+log_{10^{2}}10^{3}+log_{10^{3}}10^{4}$$&lt;/p&gt;
&lt;p&gt;$$\Rightarrow b=2+\frac{3}{2}+\frac{4}{3}=4\frac{5}{6}$$&lt;/p&gt;
&lt;p&gt;此題只要用到一些基本的對數律就可作答，答案選(3)。&lt;/p&gt;
&lt;p&gt;範圍:高中數學 數列級數 難度:★&lt;/p&gt;
&lt;p&gt;&lt;img loading="lazy" src="https://mathmanstudio.com/wp-content/uploads/2023/03/234-1024x289.png"&gt;&lt;/p&gt;
&lt;p&gt;Ans:如果此三元一次方程組無解則delta必等於0，如果delta不等於0就會有一組解，我們算一下delta的行列式值&lt;/p&gt;
&lt;p&gt;&lt;img loading="lazy" src="https://mathmanstudio.com/wp-content/uploads/2023/03/23456.png"&gt;&lt;/p&gt;
&lt;p&gt;接下來檢驗deltaX,deltaY,deltaZ的情況如果其中有一不為0的話必定無解，如果三個都等於0有可能無解或是無限多解，但這樣解有點小累&amp;hellip;.所以我把c=3代回原方程組發現，&lt;/p&gt;
&lt;p&gt;&lt;img loading="lazy" src="https://mathmanstudio.com/wp-content/uploads/2023/03/23.png"&gt;&lt;/p&gt;
&lt;p&gt;第一式和第三式相同，所以這種情況會是兩平面重合一平面穿過去為無限多解的情況，由此可見c=-2的情況必定是無解，此題答案選(2)。&lt;/p&gt;
&lt;p&gt;範圍:高中數學 三元一次聯立方程式 難度:★★&lt;/p&gt;
&lt;p&gt;&lt;img loading="lazy" src="https://mathmanstudio.com/wp-content/uploads/2023/03/25-1024x202.png"&gt;&lt;/p&gt;
&lt;p&gt;先嘗試畫個示意圖吧!!&lt;/p&gt;
&lt;p&gt;&lt;img loading="lazy" src="https://mathmanstudio.com/wp-content/uploads/2023/03/geogebra-export-1-2-1024x608.png"&gt;&lt;/p&gt;
&lt;p&gt;題目說對角線長等於1所以可以得到
$$a^{2}+b^{2}+c^{2}=1$$
再由P到Y軸的距離可得
$$a^{2}+c^{2}=\frac{2}{3}$$
，此兩式相減可得
$$b^{2}=\frac{1}{3}\Rightarrow b=\frac{1}{\sqrt{3}}$$
上述過程用的都是畢氏定理。&lt;/p&gt;
&lt;p&gt;接著就用空間中兩向量夾角的餘弦值公式即可，忘記的同學可以複習一下&lt;strong&gt;內積的定義&lt;/strong&gt;即可。&lt;/p&gt;
&lt;p&gt;我取X軸的方向向量
$$(1,0,0)$$
與OP向量
$$(a,\frac{1}{\sqrt{3}},c)$$
得到
$$cos45^{\circ}=\frac{a+0+0}{1\times \sqrt{a^{2}+\frac{1}{3}+c^{2}}}=\frac{\sqrt{2}}{2}$$
接著用
$$b=\sum_{n=1}^{3}log_{a_{n}}a_{n+1}=log_{a_{1}}a_{2}+log_{a_{2}}a_{3}+log_{a_{3}}a_{4}$$0
代回上式可得
$$b=\sum_{n=1}^{3}log_{a_{n}}a_{n+1}=log_{a_{1}}a_{2}+log_{a_{2}}a_{3}+log_{a_{3}}a_{4}$$1
現在a,b都知道了在用對角線和為1即可得到
$$b=\sum_{n=1}^{3}log_{a_{n}}a_{n+1}=log_{a_{1}}a_{2}+log_{a_{2}}a_{3}+log_{a_{3}}a_{4}$$2&lt;/p&gt;
&lt;p&gt;本題答案選(4)。&lt;/p&gt;
&lt;p&gt;範圍:高中數學 空間向量 難度:★★★&lt;/p&gt;
&lt;p&gt;&lt;img loading="lazy" src="https://mathmanstudio.com/wp-content/uploads/2023/03/254-2-1024x161.png"&gt;&lt;/p&gt;
&lt;p&gt;Ans:題目說g(x)整除f(x)所以先用長除法除除看，除了之後可以得到
$$b=\sum_{n=1}^{3}log_{a_{n}}a_{n+1}=log_{a_{1}}a_{2}+log_{a_{2}}a_{3}+log_{a_{3}}a_{4}$$3
由此兩式可以解得兩個情形
$$b=\sum_{n=1}^{3}log_{a_{n}}a_{n+1}=log_{a_{1}}a_{2}+log_{a_{2}}a_{3}+log_{a_{3}}a_{4}$$4
與
$$b=\sum_{n=1}^{3}log_{a_{n}}a_{n+1}=log_{a_{1}}a_{2}+log_{a_{2}}a_{3}+log_{a_{3}}a_{4}$$5
解著題目說g(x)=0有虛根所以&lt;code&gt;判別式必須要小於&lt;/code&gt;&lt;strong&gt;0&lt;/strong&gt;，
$$b=\sum_{n=1}^{3}log_{a_{n}}a_{n+1}=log_{a_{1}}a_{2}+log_{a_{2}}a_{3}+log_{a_{3}}a_{4}$$6
是合理的因此
$$b=\sum_{n=1}^{3}log_{a_{n}}a_{n+1}=log_{a_{1}}a_{2}+log_{a_{2}}a_{3}+log_{a_{3}}a_{4}$$7
是正解，最後將a,k套回原本的除法母式可得
$$b=\sum_{n=1}^{3}log_{a_{n}}a_{n+1}=log_{a_{1}}a_{2}+log_{a_{2}}a_{3}+log_{a_{3}}a_{4}$$8
解得
$$b=\sum_{n=1}^{3}log_{a_{n}}a_{n+1}=log_{a_{1}}a_{2}+log_{a_{2}}a_{3}+log_{a_{3}}a_{4}$$9
所以本題答案選(1)(4)。&lt;/p&gt;
&lt;p&gt;範圍:高中數學 多項式 難度:★★&lt;/p&gt;</description></item></channel></rss>