《2026年步步高精准讲练高中数学选择性必修第二册人教版》

已知数列$\{ a_{n}\}$满足$a_{1}=1$，$a_{n + 1}=2a_{n}+1(n\in \mathbf{N}^{*})$.
(1)记$b_{n}=a_{n}+1$，证明：数列$\{ b_{n}\}$为等比数列，并求数列$\{ b_{n}\}$的通项公式；
(2)记数列$\{ b_{n}\}$的前$n$项和为$T_{n}$，证明：$\dfrac {b_{2}}{T_{1}T_{2}}+\dfrac {b_{3}}{T_{2}T_{3}}+·s +\dfrac {b_{n + 1}}{T_{n}T_{n + 1}}＜\dfrac {1}{2}$.

答案：
(1)解 由$a_{n+1}=2a_{n}+1$，可得$a_{n+1}+1=2(a_{n}+1)$，所以$\{b_{n}\}$是首项为$a_{1}+1=2$，公比为$2$的等比数列，所以$b_{n}=a_{n}+1=2^{n}$．
(2)证明 易得$T_{n}=\frac{2×(1-2^{n})}{1-2}=2(2^{n}-1)$，于是$\frac{b_{n+1}}{T_{n}T_{n+1}}=\frac{T_{n+1}-T_{n}}{T_{n}T_{n+1}}=\frac{1}{T_{n}}-\frac{1}{T_{n+1}}=\frac{1}{2}\left[\left(\frac{1}{2^{2}-1}-\frac{1}{2^{2}-1}\right)+\left(\frac{1}{2^{2}-1}-\frac{1}{2^{3}-1}\right)+·s+\left(\frac{1}{2^{n}-1}-\frac{1}{2^{n+1}-1}\right)\right]=\frac{1}{2}\left(1-\frac{1}{2^{n+1}-1}\right)$，因为$\frac{1}{2^{n+1}-1}＞0$，所以$\frac{b_{2}}{T_{1}T_{2}}+\frac{b_{3}}{T_{2}T_{3}}+·s+\frac{b_{n+1}}{T_{n}T_{n+1}}=\frac{1}{2}\left(1-\frac{1}{2^{n+1}-1}\right)＜\frac{1}{2}$．

例 1 已知数列 $ a_{n}=(-1)^{n}n $，求数列$\{ a_{n}\}$的前 $ n $ 项和 $ S_{n} $.
反思感悟：

答案： 解 方法一 （并项求和）若$n$是偶数，则$S_{n}=(-1 + 2) + (-3 + 4)+ (-5 + 6) + ·s + [-(n - 1) + n] = \frac{n}{2}$．若$n$是奇数，则$S_{n}=(-1 + 2) + (-3 + 4) + (-5 + 6) + ·s + (-n)= \frac{n - 1}{2} - n = -\frac{n + 1}{2}$．综上所述，$S_{n} = \begin{cases} \frac{n}{2}, n 为偶数, \\ -\frac{n + 1}{2}, n 为奇数. \end{cases}$方法二 （分组求和）若$n$是偶数，则$S_{n} = -1 + 2 - 3 + 4 - ·s - (n - 1) + n = -[1 + 3 + ·s + (n - 1)] + (2 + 4 + ·s + n) = -\frac{n^{2}}{4} + \frac{n(n + 2)}{4} = \frac{n}{2}$；若$n$是奇数，则$S_{n} = -1 + 2 - 3 + 4 - ·s + (n - 1) - n = -(1 + 3 + ·s + n) + [2 + 4 + ·s + (n - 1)] = -\frac{(n + 1)^{2}}{4} + \frac{(n - 1)(n + 1)}{4} = -\frac{n + 1}{2}$，综上所述，$S_{n} = \begin{cases} \frac{n}{2}, n 为偶数, \\ -\frac{n + 1}{2}, n 为奇数. \end{cases}$

跟踪训练 1 已知数列$\{ a_{n}\}$满足$ a_{n}=(-1)^{n}n^{2} $，则$ a_{1}+a_{2}+a_{3}+·s +a_{2n+1} $等于(A)

A.$ -(n+1)(2n+1) $
B.$ (n+1)(2n+1) $
C.$ -n(n+1) $
D.$ n(n+1) $

答案： $[a_{1} + a_{2} + a_{3} + ·s + a_{2n + 1} = -1^{2} + 2^{2} - 3^{2} + 4^{2} - 5^{2} + ·s + (2n)^{2} - (2n + 1)^{2} = -1 + (2^{2} - 3^{2}) + (4^{2} - 5^{2}) + ·s + [(2n)^{2} - (2n + 1)^{2}] = -1 - (2 + 3) - (4 + 5) - ·s - (2n + 2n + 1) = -[1 + 2 + 3 + 4 + 5 + ·s + (2n + 1)] = -\frac{(2n + 1)(2n + 2)}{2} = -(n + 1)(2n + 1)$．