《2025年成才之路高中新课程学习指导高中数学选择性必修第二册人教版》

▶跟踪训练$3$
化简$S_{n} = n + (n - 1) × 2 + (n - 2) × 2^{2} + ·s + 2 × 2^{n - 2} + 2^{n - 1}$的结果是 (D)

A.$2^{n + 1} + n - 2$
B.$2^{n + 1} - n + 2$
C.$2^{n} - n - 2$
D.$2^{n + 1} - n - 2$

答案： 3.D $S_{n}=n+(n-1)×2+(n-2)×2^{2}+·s +2×2^{n-2}+2^{n-1}$, ①
$2S_{n}=n×2+(n-1)×2^{2}+(n-2)×2^{3}+·s +2×2^{n-1}+2^{n}$,②
②$-$①得,$S_{n}=-n+2+2^{2}+·s +2^{n-1}+2^{n}=-n+\frac {2(1-2^{n})}{1-2}$
$=-n+2^{n+1}-2=2^{n+1}-n-2$.

例 4. 已知数列$\{ a_{n}\}$满足$a_{1} = 1,\frac{a_{n + 1}}{a_{n}} = \frac{n}{n + 2}$，求$\{ a_{n}\}$的前$6$项和.
▶[方法总结$4$]
[方法总结$4$]
常见的裂项的形式：
(1)$\frac{1}{n(n + 1)} = \frac{1}{n} -$
$\frac{1}{n + 1},\frac{1}{n(n + k)} =$
$\frac{1}{k}\left( \frac{1}{n} - \frac{1}{n + k} \right)$；
(2)$\frac{1}{\sqrt{n + 1} + \sqrt{n}} =$
$\sqrt{n + 1} - \sqrt{n}$，
$\frac{1}{\sqrt{n + k} + \sqrt{n}} =$
$\frac{1}{k}(\sqrt{n + k} - \sqrt{n})$；
(3)$\ln\left( 1 + \frac{1}{n} \right) = \ln(n$
$+ 1) - \ln n$，
$\ln\left( 1 - \frac{1}{n} \right) = \ln(n -$
$1) - \ln n(n \geq 2)$；
(4)若$\{ a_{n}\}$是等差数列，公差为$d$，则
$\frac{1}{a_{n}a_{n + 1}} = \frac{1}{d}\left( \frac{1}{a_{n}} - \right.$
$\left. \frac{1}{a_{n + 1}} \right)$
如$\frac{1}{(2n - 1)(2n + 1)} =$
$\frac{1}{2}\left( \frac{1}{2n - 1} - \right.$
$\left. \frac{1}{2n + 1} \right)$；
(5)$\frac{1}{n(n + 1)(n + 2)} =$
$\frac{1}{2}\left\lbrack \frac{1}{n(n + 1)} - \right.$
$\left. \frac{1}{(n + 1)(n + 2)} \right\rbrack$；
(6)$\frac{2^{n}}{(2^{n} + 1)(2^{n + 1} + 1)} =$
$\frac{1}{2^{n} + 1} - \frac{1}{2^{n + 1} + 1}$；
(7)$\frac{n + 1}{n^{2}(n + 2)^{2}} =$
$\frac{1}{4}\left\lbrack \frac{1}{n^{2}} - \frac{1}{(n + 2)^{2}} \right\rbrack$.

答案： 例4:【解析】 步骤一:求数列$\{ a_{n}\}$的通项公式
已知$\frac {a_{n+1}}{a_{n}}=\frac {n}{n+2}$,我们利用累乘法来求数列的通项公式.
$a_{n}=\frac {a_{n}}{a_{n-1}}·\frac {a_{n-1}}{a_{n-2}}··s·\frac {a_{2}}{a_{1}}· a_{1}$
$=\frac {n-1}{n+1}·\frac {n-2}{n}·\frac {n-3}{n-1}··s·\frac {2}{4}·\frac {1}{3}·1$
$=\frac {2}{n(n+1)}$
当$n=1$时,$a_{1}=\frac {2}{1×(1+1)}=1$,满足通项公式,
所以$a_{n}=\frac {2}{n(n+1)}$.
步骤二:求数列$\{ a_{n}\}$的前6项和
对$a_{n}=\frac {2}{n(n+1)}$进行裂项,可得$a_{n}=2\left(\frac {1}{n}-\frac {1}{n+1}\right)$.
设数列$\{ a_{n}\}$的前$n$项和为$S_{n}$,则:
$S_{6}=2\left[\left(1-\frac {1}{2}\right)+\left(\frac {1}{2}-\frac {1}{3}\right)+\left(\frac {1}{3}-\frac {1}{4}\right)+\left(\frac {1}{4}-\frac {1}{5}\right)+\left(\frac {1}{5}-\frac {1}{6}\right)+\left(\frac {1}{6}-\frac {1}{7}\right)\right]$
$=2\left(1-\frac {1}{7}\right)$
$=\frac {12}{7}$.