請教單選第三題

請教單選第五題

請教單選第六題

請教複選第12題

填充題1
設級數\(f(n)=1^n-2^n+3^n-4^n+\ldots+2015^n-2016^n+2017^n\)，求\(\displaystyle \frac{f(1)f(2)}{f(3)}=\)　　　。
[解答]
\(f(3)=1^3-2^3+3^3-4^3-\dots-2016^3+2017^3\)
\(=(1^3+2^3+3^3+4^3+\dots+2017^3)-2(2^3+4^3+\dots+2016^3)\)
\(\displaystyle=\left(\frac{2017\times2018}{2}\right)^2-2^4(1^3+2^3+\dots+1008^3)\)
\(\displaystyle=2017^2\times1009^2-16\times\left(\frac{1008\times1009}{2}\right)^2\)
\(=2017^2\times1009^2-4\times1008^2\times1009^2\)
\(=1009^2(2017^2-2016^2)\)
\(=1009^2(2017-2016)(2017+2016)\)
\(=1009^2(1)(4033)\)
\(=1009^2\times4033\)
\(\displaystyle \frac{f(1)f(2)}{f(3)}=\frac{1009^2\times2017}{1009^2\times4033}=\frac{2017}{4033}\)

填充題5
求值：\(\displaystyle \sum_{k=1}^{\infty}\frac{1}{k^3+8k^2+15k}=\)　　　。
[解答]
\(\displaystyle \sum_{k=1}^{\infty} \frac{1}{k^3 + 8k^2 + 15k}\)

\(\displaystyle= \sum_{k=1}^{\infty} \frac{1}{k(k^2 + 8k + 15)}\)

\(\displaystyle= \sum_{k=1}^{\infty} \frac{1}{k(k+3)(k+5)}\)

\(\displaystyle= \frac{1}{15} \sum_{k=1}^{\infty} \left(\frac{1}{k} - \frac{1}{k+3}\right) - \frac{1}{10} \sum_{k=1}^{\infty} \left(\frac{1}{k+3} - \frac{1}{k+5}\right)\)

\(\displaystyle= \frac{1}{15} \left[\left(1 - \frac{1}{4}\right) + \left(\frac{1}{2} - \frac{1}{5}\right) + \left(\frac{1}{3} - \frac{1}{6}\right) + \left(\frac{1}{4} - \frac{1}{7}\right) + \dots\right]\)
\(\displaystyle- \frac{1}{10} \left[\left(\frac{1}{4} - \frac{1}{6}\right) + \left(\frac{1}{5} - \frac{1}{7}\right) + \left(\frac{1}{6} - \frac{1}{8}\right) + \left(\frac{1}{7} - \frac{1}{9}\right) + \dots\right]\)

\(\displaystyle= \frac{1}{15} \left(1 + \frac{1}{2} + \frac{1}{3}\right) - \frac{1}{10} \left(\frac{1}{4} + \frac{1}{5}\right)\)

\(\displaystyle= \frac{1}{15} \times \frac{11}{6} - \frac{1}{10} \times \frac{9}{20}\)

\(\displaystyle= \frac{11}{90} - \frac{9}{200}\)

\(\displaystyle= \frac{11 \times 20 - 9 \times 9}{1800}\)

\(\displaystyle= \frac{220 - 81}{1800}\)

\(\displaystyle= \frac{139}{1800}\)

請教填充題6詳細作法版上解答和連結實在看不懂

請教填充題9

填充9
設\(P\)為直線\(x-y+5=0\)上一點，\(Q\)為雙曲線一支\(\Gamma\)：\(\displaystyle x=\sqrt{\frac{9}{4}y^2+9}\)上一點，求\(\overline{PQ}\)最小值=　　　。
[解答]
\(y=x+5\)

\(m=1\)

\(x=\sqrt{\frac{9}{4}y^2+9}>0\)(右支)

\(\Rightarrow\)整理\(\displaystyle \frac{x^2}{9}-\frac{y^2}{4}=1\)

\(a=3\),\(b=2\)

雙曲線微分切線公式:
\(y=mx\pm\sqrt{a^2m^2-b^2}\)

\(m=1\)時
\(y=x\pm\sqrt{3^2\cdot1^2-2^2}\)
\(=x\pm\sqrt{9-4}\)
\(=x\pm\sqrt{5}\)(正不合)
右支

\(\{M:y=x-\sqrt{5}\Rightarrow x-y=\sqrt{5}\)
\(\{L:y=x+5\Rightarrow x-y=-5\)

\(\displaystyle d(L,M)=\frac{|\sqrt{5}-(-5)|}{\sqrt{1^2+(-1)^2}}\)
\(\displaystyle =\frac{\sqrt{5}+5}{\sqrt{2}}\)
\(\displaystyle =\frac{(\sqrt{5}+5)\sqrt{2}}{2}\)
\(\displaystyle =\frac{\sqrt{10}+5\sqrt{2}}{2}\)

感謝各位老師幫忙練習過一遍