这篇博客探讨了今年高考中的一道数学难题，涉及数列的通项公式推导。通过巧妙的数学变换，博主展示了如何从递推关系一步步求得数列的通项，并利用裂项相消法证明了数列极限的存在性，证明过程严谨，适合数学爱好者学习参考。
听说今年的高考数学题很难，今天有空看了几题，以17题练个手。廉颇老矣，尚能饭否？！
(1)
Cn=SnanC1=a1a1=1Cn=1+(n−1)13=n+23Sn=n+23anSn−1=n+13an−1Sn−Sn−1=n+23an−n+13an−1an=n+23an−n+13an−13an=(n+2)an−(n+1)an−1(n−1)an=(n+1)an−1anan−1=n+1n−1an−1an−2=nn−2…a2a1=31anan−1⋅an−1an−2⋅a2a1=n+1n−1⋅nn−2⋯31an=(n+1)!n!2n=n(n+1)2 \begin{aligned} C_n &= \frac{S_n}{a_n} \\ C_1 &= \frac{a_1}{a_1} = 1 \\ C_n &= 1+(n-1) \frac{1}{3} = \frac{n+2}{3} \\ S_n &= \frac{n+2}{3} a_n \\ S_{n-1} &= \frac{n+1}{3} a_{n-1} \\ S_n - S_{n-1} &= \frac{n+2}{3} a_n - \frac{n+1}{3} a_{n-1} \\ a_n &= \frac{n+2}{3} a_n - \frac{n+1}{3} a_{n-1} \\ 3a_n &= (n+2)a_n - (n+1)a_{n-1} \\ (n-1)a_{n} &= (n+1)a_{n-1} \\ \frac{a_{n}}{a_{n-1}} &= \frac{n+1}{n-1} \\ \frac{a_{n-1}}{a_{n-2}} &= \frac{n}{n-2} \\ \dots \\ \frac{a_{2}}{a_{1}} &= \frac{3}{1} \\ \frac{a_{n}}{a_{n-1}} \cdot \frac{a_{n-1}}{a_{n-2}} \cdot \frac{a_{2}}{a_{1}} &= \frac{n+1}{n-1} \cdot \frac{n}{n-2} \cdots \frac{3}{1} \\ a_n &= \frac{\frac{(n+1)!}{n!}}{\frac{2}{n}} \\ &= \frac{n(n+1)}{2} \end{aligned} CnC1CnSnSn−1Sn−Sn−1an3an(n−1)anan−1anan−2an−1…a1a2an−1an⋅an−2an−1⋅a1a2an=anSn=a1a1=1=1+(n−1)31=3n+2=3n+2an=3n+1an−1=3n+2an−3n+1an−1=3n+2an−3n+1an−1=(n+2)an−(n+1)an−1=(n+1)an−1=n−1n+1=n−2n=13=n−1n+1⋅n−2n⋯13=n2n!(n+1)!=2n(n+1)
(2)
1an=2n(n+1)=2(1n−1n+1)1a1+⋯+1an=2(1−12+12−13+⋯+1n−1n+1)<2 \begin{aligned} \frac{1}{a_n} &= \frac{2}{n(n+1)} \\ &= 2(\frac{1}{n} - \frac{1}{n+1}) \\ \frac{1}{a_1} + \cdots + \frac{1}{a_n} &= 2(1-\frac{1}{2} + \frac{1}{2}-\frac{1}{3} + \cdots + \frac{1}{n} - \frac{1}{n+1}) < 2 \end{aligned} an1a11+⋯+an1=n(n+1)2=2(n1−n+11)=2(1−21+21−31+⋯+n1−n+11)<2
扫一扫
被折叠的 条评论
为什么被折叠?
到【灌水乐园】发言
