回覆 5# Superconan 的帖子
以下是整理本串有的解答以及自己算的部分解答
因為久沒算題目，可能錯誤百出，還請各位不吝指教
1. (9,9,6)
2.\(\left[
\begin{array}{cc}
2 & -3 \\
3 & -4 \\
\end{array}
\right] \)
3.(1) \(\displaystyle 2\sqrt{2}\)
(2) 1，\(\displaystyle \frac{7}{8}\)
5(1) \(\displaystyle h=\frac{-b}{3a}, p=c-\frac{b^2}{3a} , k=d+\frac{2b^3}{27a^2}-\frac{bc}{3a}\)
7. \(\displaystyle \Sigma_{k=1}^n \displaystyle k^4 =\frac{1}{5}n^5+\frac{1}{2}n^4+\frac{1}{3}n^3-\frac{1}{30}n\)
8(1). -1 (2)\(\displaystyle \sqrt{11}i\)
9.(1) \(\displaystyle
\left\{
\begin{array}{**lr**}
a_1=0 &\\
a_n= 2a_{n-1}+(4n-6) , n \geq 2&
\end{array}
\right. \)
(2) \(\displaystyle
\left\{
\begin{array}{**lr**}
a_1=0,&\\
a_n= 3\times 2^n-4n-2 , n \geq 2&
\end{array}
\right. \)
10.x=4
11. 2
12. \(\displaystyle 2(1+\sqrt{3}) \)
[ 本帖最後由 satsuki931000 於 2024-4-23 10:57 編輯 ]
