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2025年练习生高中数学选择性必修第二册人教版
注：目前有些书本章节名称可能整理的还不是很完善，但都是按照顺序排列的，请同学们按照顺序仔细查找。练习册 2025年练习生高中数学选择性必修第二册人教版 答案主要是用来给同学们做完题方便对答案用的，请勿直接抄袭。
15. [2023·新课标Ⅰ卷]设等差数列$\{ a_{n}\}$的公差为$d$,且$d＞1$.令$b_{n}=\frac {n^{2}+n}{a_{n}}$,记$S_{n},T_{n}$分别为数列$\{ a_{n}\},\{ b_{n}\}$的前$n$项和.
(1)若$3a_{2}=3a_{1}+a_{3},S_{3}+T_{3}=21$,求$\{ a_{n}\}$的通项公式;
(2)若$\{ b_{n}\}$为等差数列,且$S_{99}-T_{99}=99$,求$d$.
(1)若$3a_{2}=3a_{1}+a_{3},S_{3}+T_{3}=21$,求$\{ a_{n}\}$的通项公式;
(2)若$\{ b_{n}\}$为等差数列,且$S_{99}-T_{99}=99$,求$d$.
答案：
15.【解】
(1)$\because3a_{2}=3a_{1}+a_{3},\therefore3d=a_{1}+2d,\therefore a_{1}=d$,则$a_{n}=nd(d＞1)$,
$\therefore b_{n}=\frac{n + 1}{d}$.
$\therefore S_{3}=a_{1}+a_{2}+a_{3}=6d,T_{3}=b_{1}+b_{2}+b_{3}=\frac{9}{d}$,
$\therefore6d+\frac{9}{d}=21$,整理,得$2d^{2}-7d + 3 = 0$,
即$(2d - 1)(d - 3)=0$,解得$d = 3$或$d=\frac{1}{2}$(舍去).
$\therefore a_{n}=3n,n\in\mathbf{N}^*$.
(2)方法一若$\{ b_{n}\}$为等差数列,
则$b_{1}+b_{3}=2b_{2}$,即$\frac{2}{a_{1}}+\frac{12}{a_{3}}=2·\frac{6}{a_{2}}$
整理,得$a_{1}^{2}-3da_{1}+2d^{2}=0$.
解得$a_{1}=d$或$a_{1}=2d$.
当$a_{1}=d$时,$a_{n}=dn,b_{n}=\frac{n^{2}+n}{dn}=\frac{n + 1}{d}$,
$\therefore S_{99}-T_{99}=\frac{99}{2}(d + 99d)-\frac{99}{2}(\frac{2}{d}+\frac{100}{d})=99$.
整理,得$50d^{2}-d - 51 = 0$,
解得$d=\frac{51}{50}$或$d=-1$(舍去).
当$a_{1}=2d$时,$a_{n}=d(n + 1),b_{n}=\frac{n^{2}+n}{d(n + 1)}=\frac{n}{d}$,
$\therefore S_{99}-T_{99}=\frac{99}{2}(2d + 100d)-\frac{99}{2}(\frac{2}{d}+\frac{99}{d})=99$.
整理,得$51d^{2}-d - 50 = 0$,
解得$d=-\frac{50}{51}$或$d = 1$.
$\because d＞1,\therefore$此时无解.
综上所述,$d=\frac{51}{50}$.
方法二设等差数列$\{ b_{n}\}$的公差为$r,b_{0}=b_{1}-r,a_{0}=a_{1}-d$,
则由$b_{n}=\frac{n^{2}+n}{a_{n}}$,得$b_{0}+nr=\frac{n^{2}+n}{a_{0}+nd}$,
$\therefore n^{2}+n=(a_{0}+nd)(b_{0}+nr)=drn^{2}+(db_{0}+ra_{0})n+a_{0}b_{0}$,
$\therefore\begin{cases}dr = 1,\\db_{0}+ra_{0}=1,\\a_{0}b_{0}=0.\end{cases}(d＞1,0＜r＜1)$.
$\therefore S_{99}-T_{99}=\sum_{n = 1}^{99}(a_{n}-b_{n})=\sum_{n = 1}^{99}\frac{99×100}{2}×(d - r)=99$,
$\therefore a_{0}-b_{0}+50(d - r)=1$.
①当$a_{0}=0$时,$db_{0}=1,dr = 1,50(d - r)=1 + b_{0}$,
则$\frac{1}{50}(d-\frac{1}{d})=1+\frac{1}{d}$,
$\therefore50d^{2}-51d - 50 = 0$,即$(50d - 51)(d + 1)=0$,
解得$d=\frac{51}{50}$或$d=-1$(舍去).
②当$b_{0}=0$时,$ra_{0}=1,dr = 1,a_{0}+50(d - r)=1$,
则$\frac{1}{50}+\frac{1}{r}(d-\frac{1}{50})=1$,
$\therefore50r^{2}+r - 51 = 0$,即$(50r + 51)(r - 1)=0$,
解得$r=-\frac{51}{50}$(舍去)或$r = 1$(舍去).
此时不存在满足题意的$d$.
综上所述,$d=\frac{51}{50}$.
(1)$\because3a_{2}=3a_{1}+a_{3},\therefore3d=a_{1}+2d,\therefore a_{1}=d$,则$a_{n}=nd(d＞1)$,
$\therefore b_{n}=\frac{n + 1}{d}$.
$\therefore S_{3}=a_{1}+a_{2}+a_{3}=6d,T_{3}=b_{1}+b_{2}+b_{3}=\frac{9}{d}$,
$\therefore6d+\frac{9}{d}=21$,整理,得$2d^{2}-7d + 3 = 0$,
即$(2d - 1)(d - 3)=0$,解得$d = 3$或$d=\frac{1}{2}$(舍去).
$\therefore a_{n}=3n,n\in\mathbf{N}^*$.
(2)方法一若$\{ b_{n}\}$为等差数列,
则$b_{1}+b_{3}=2b_{2}$,即$\frac{2}{a_{1}}+\frac{12}{a_{3}}=2·\frac{6}{a_{2}}$
整理,得$a_{1}^{2}-3da_{1}+2d^{2}=0$.
解得$a_{1}=d$或$a_{1}=2d$.
当$a_{1}=d$时,$a_{n}=dn,b_{n}=\frac{n^{2}+n}{dn}=\frac{n + 1}{d}$,
$\therefore S_{99}-T_{99}=\frac{99}{2}(d + 99d)-\frac{99}{2}(\frac{2}{d}+\frac{100}{d})=99$.
整理,得$50d^{2}-d - 51 = 0$,
解得$d=\frac{51}{50}$或$d=-1$(舍去).
当$a_{1}=2d$时,$a_{n}=d(n + 1),b_{n}=\frac{n^{2}+n}{d(n + 1)}=\frac{n}{d}$,
$\therefore S_{99}-T_{99}=\frac{99}{2}(2d + 100d)-\frac{99}{2}(\frac{2}{d}+\frac{99}{d})=99$.
整理,得$51d^{2}-d - 50 = 0$,
解得$d=-\frac{50}{51}$或$d = 1$.
$\because d＞1,\therefore$此时无解.
综上所述,$d=\frac{51}{50}$.
方法二设等差数列$\{ b_{n}\}$的公差为$r,b_{0}=b_{1}-r,a_{0}=a_{1}-d$,
则由$b_{n}=\frac{n^{2}+n}{a_{n}}$,得$b_{0}+nr=\frac{n^{2}+n}{a_{0}+nd}$,
$\therefore n^{2}+n=(a_{0}+nd)(b_{0}+nr)=drn^{2}+(db_{0}+ra_{0})n+a_{0}b_{0}$,
$\therefore\begin{cases}dr = 1,\\db_{0}+ra_{0}=1,\\a_{0}b_{0}=0.\end{cases}(d＞1,0＜r＜1)$.
$\therefore S_{99}-T_{99}=\sum_{n = 1}^{99}(a_{n}-b_{n})=\sum_{n = 1}^{99}\frac{99×100}{2}×(d - r)=99$,
$\therefore a_{0}-b_{0}+50(d - r)=1$.
①当$a_{0}=0$时,$db_{0}=1,dr = 1,50(d - r)=1 + b_{0}$,
则$\frac{1}{50}(d-\frac{1}{d})=1+\frac{1}{d}$,
$\therefore50d^{2}-51d - 50 = 0$,即$(50d - 51)(d + 1)=0$,
解得$d=\frac{51}{50}$或$d=-1$(舍去).
②当$b_{0}=0$时,$ra_{0}=1,dr = 1,a_{0}+50(d - r)=1$,
则$\frac{1}{50}+\frac{1}{r}(d-\frac{1}{50})=1$,
$\therefore50r^{2}+r - 51 = 0$,即$(50r + 51)(r - 1)=0$,
解得$r=-\frac{51}{50}$(舍去)或$r = 1$(舍去).
此时不存在满足题意的$d$.
综上所述,$d=\frac{51}{50}$.
16. [2023·新课标Ⅱ卷]已知$\{ a_{n}\}$为等差数列,$b_{n}=$$\left\{\begin{array}{l} a_{n}-6,n为奇数,\\ 2a_{n},n为偶数.\end{array}\right.$记$S_{n},T_{n}$分别为数列$\{ a_{n}\},$$\{ b_{n}\}$的前$n$项和,若$S_{4}=32,T_{3}=16$.
(1)求$\{ a_{n}\}$的通项公式;
(2)证明:当$n＞5$时,$T_{n}＞S_{n}$.
(1)求$\{ a_{n}\}$的通项公式;
(2)证明:当$n＞5$时,$T_{n}＞S_{n}$.
答案：
16.
(1)【解】设等差数列$\{ a_{n}\}$的公差为$d$.
因为$b_{n}=\begin{cases}a_{n}-6,n为奇数,\\2a_{n},n为偶数,\end{cases}$
且$S_{4}=32,T_{3}=16$,
所以$\begin{cases}4a_{1}+6d=32,\\a_{1}-6 + 2a_{1}+2d-6 + a_{1}+2d-6=16,\end{cases}$
解得$\begin{cases}d = 2,\\a_{1}=5,\end{cases}$
所以$a_{n}=5 + 2(n - 1)=2n + 3$.
(2)【证明】由
(1)可知,$S_{n}=\frac{n(a_{1}+a_{n})}{2}=\frac{n(5 + 2n + 3)}{2}=\frac{n^{2}+4n}{2}$.
若$n$为偶数,则
$T_{n}=(b_{1}+b_{3}+·s+b_{n - 1})+(b_{2}+b_{4}+·s+b_{n})$
$=(a_{1}-6 + a_{3}-6+·s+a_{n - 1}-6)+2(a_{2}+a_{4}+·s+a_{n})$
$=(5 + 9+·s+2n - 3 - 6)+2(7 + 11+·s+2n + 3)$
$=\frac{(5 + 2n + 1)×\frac{n}{2}}{2}-3n + 2×\frac{(7 + 2n + 3)×\frac{n}{2}}{2}$
$=\frac{3}{2}n^{2}+\frac{7}{2}n - 3n - (n^{2}+4n)$
$=\frac{1}{2}n^{2}-\frac{1}{2}n=\frac{1}{2}n(n - 1)＞0$,
即$T_{n}＞S_{n}$.
若$n$为奇数,则$n - 1$为偶数,所以
$T_{n}=T_{n - 1}+b_{n}$
$=\frac{3}{2}(n - 1)^{2}+\frac{5}{2}(n - 1)+2n + 3 - 6$
$=\frac{3}{2}n^{2}+\frac{5}{2}n - 5$
所以当$n＞5$时,
$T_{n}-S_{n}=\frac{3}{2}n^{2}+\frac{5}{2}n - 5-(n^{2}+4n)$
$=\frac{1}{2}n^{2}-\frac{3}{2}n - 5$
$=\frac{1}{2}(n^{2}-3n - 10)$
$=\frac{1}{2}(n + 2)(n - 5)＞0$,
即$T_{n}＞S_{n}$.
综上所述,当$n＞5$时,$T_{n}＞S_{n}$.
(1)【解】设等差数列$\{ a_{n}\}$的公差为$d$.
因为$b_{n}=\begin{cases}a_{n}-6,n为奇数,\\2a_{n},n为偶数,\end{cases}$
且$S_{4}=32,T_{3}=16$,
所以$\begin{cases}4a_{1}+6d=32,\\a_{1}-6 + 2a_{1}+2d-6 + a_{1}+2d-6=16,\end{cases}$
解得$\begin{cases}d = 2,\\a_{1}=5,\end{cases}$
所以$a_{n}=5 + 2(n - 1)=2n + 3$.
(2)【证明】由
(1)可知,$S_{n}=\frac{n(a_{1}+a_{n})}{2}=\frac{n(5 + 2n + 3)}{2}=\frac{n^{2}+4n}{2}$.
若$n$为偶数,则
$T_{n}=(b_{1}+b_{3}+·s+b_{n - 1})+(b_{2}+b_{4}+·s+b_{n})$
$=(a_{1}-6 + a_{3}-6+·s+a_{n - 1}-6)+2(a_{2}+a_{4}+·s+a_{n})$
$=(5 + 9+·s+2n - 3 - 6)+2(7 + 11+·s+2n + 3)$
$=\frac{(5 + 2n + 1)×\frac{n}{2}}{2}-3n + 2×\frac{(7 + 2n + 3)×\frac{n}{2}}{2}$
$=\frac{3}{2}n^{2}+\frac{7}{2}n - 3n - (n^{2}+4n)$
$=\frac{1}{2}n^{2}-\frac{1}{2}n=\frac{1}{2}n(n - 1)＞0$,
即$T_{n}＞S_{n}$.
若$n$为奇数,则$n - 1$为偶数,所以
$T_{n}=T_{n - 1}+b_{n}$
$=\frac{3}{2}(n - 1)^{2}+\frac{5}{2}(n - 1)+2n + 3 - 6$
$=\frac{3}{2}n^{2}+\frac{5}{2}n - 5$
所以当$n＞5$时,
$T_{n}-S_{n}=\frac{3}{2}n^{2}+\frac{5}{2}n - 5-(n^{2}+4n)$
$=\frac{1}{2}n^{2}-\frac{3}{2}n - 5$
$=\frac{1}{2}(n^{2}-3n - 10)$
$=\frac{1}{2}(n + 2)(n - 5)＞0$,
即$T_{n}＞S_{n}$.
综上所述,当$n＞5$时,$T_{n}＞S_{n}$.
17. [2022·浙江高考]已知等差数列$\{ a_{n}\}$的首项$a_{1}=-1$，公差$d ＞ 1$。记$\{ a_{n}\}$的前$n$项和为$S_{n}(n\in\mathbf{N}^{*})$。
(1)若$S_{4}-2a_{2}a_{3}+6 = 0$，求$S_{n}$；
(2)若对于每个$n\in\mathbf{N}^{*}$，存在实数$c_{n}$，使$a_{n}+c_{n}$，$a_{n + 1}+4c_{n}$，$a_{n + 2}+15c_{n}$成等比数列，求$d$的取值范围。
(1)若$S_{4}-2a_{2}a_{3}+6 = 0$，求$S_{n}$；
(2)若对于每个$n\in\mathbf{N}^{*}$，存在实数$c_{n}$，使$a_{n}+c_{n}$，$a_{n + 1}+4c_{n}$，$a_{n + 2}+15c_{n}$成等比数列，求$d$的取值范围。
答案：
17.【解】
(1)设$a_{n}=(n - 1)d - 1$.
依题意,得$6d - 4 - 2(d - 1)(2d - 1)+6 = 0$,
解得$d = 3$或$d = 0$(舍去).
所以$a_{n}=3n - 4,n\in\mathbf{N}^*$.
于是$S_{n}=\frac{n(a_{1}+a_{n})}{2}=\frac{n(3n - 5)}{2},n\in\mathbf{N}^*$.
(2)设$a_{n}=(n - 1)d - 1$.
依题意,得$[c_{n}+(n - 1)d - 1][15c_{n}+(n + 1)d - 1]=(4c_{n}+nd - 1)^{2}$,
所以$15c_{n}^{2}+[(16n - 14)d - 16]c_{n}+n^{2}d^{2}-2nd + 1=16c_{n}^{2}+8(nd - 1)c_{n}+n^{2}d^{2}-2nd + 1$,
整理,得$c_{n}^{2}+[(14 - 8n)d + 8]c_{n}+d^{2}=0$,
故$\Delta=[(14 - 8n)d + 8]^{2}-4d^{2}\geq0$,即$[(12 - 8n)d + 8][(16 - 8n)d + 8]\geq0$,
即$[(3 - 2n)d + 2][(2 - n)d + 1]\geq0$对任意正整数$n$成立.
当$n = 1$时,显然成立;
当$n = 2$时,$-d + 2\geq0$,则$d\leq2$;
当$n\geq3$时,$[(2n - 3)d - 2][(n - 2)d - 1]＞(2n - 5)(n - 3)\geq0$.
综上所述,$1＜d\leq2$.
(1)设$a_{n}=(n - 1)d - 1$.
依题意,得$6d - 4 - 2(d - 1)(2d - 1)+6 = 0$,
解得$d = 3$或$d = 0$(舍去).
所以$a_{n}=3n - 4,n\in\mathbf{N}^*$.
于是$S_{n}=\frac{n(a_{1}+a_{n})}{2}=\frac{n(3n - 5)}{2},n\in\mathbf{N}^*$.
(2)设$a_{n}=(n - 1)d - 1$.
依题意,得$[c_{n}+(n - 1)d - 1][15c_{n}+(n + 1)d - 1]=(4c_{n}+nd - 1)^{2}$,
所以$15c_{n}^{2}+[(16n - 14)d - 16]c_{n}+n^{2}d^{2}-2nd + 1=16c_{n}^{2}+8(nd - 1)c_{n}+n^{2}d^{2}-2nd + 1$,
整理,得$c_{n}^{2}+[(14 - 8n)d + 8]c_{n}+d^{2}=0$,
故$\Delta=[(14 - 8n)d + 8]^{2}-4d^{2}\geq0$,即$[(12 - 8n)d + 8][(16 - 8n)d + 8]\geq0$,
即$[(3 - 2n)d + 2][(2 - n)d + 1]\geq0$对任意正整数$n$成立.
当$n = 1$时,显然成立;
当$n = 2$时,$-d + 2\geq0$,则$d\leq2$;
当$n\geq3$时,$[(2n - 3)d - 2][(n - 2)d - 1]＞(2n - 5)(n - 3)\geq0$.
综上所述,$1＜d\leq2$.
18. [2021·新高考Ⅰ卷]已知数列$\{ a_{n}\}$满足$a_{1}=1$，
$a_{n + 1}=\begin{cases}a_{n}+1,n为奇数,\\a_{n}+2,n为偶数.\end{cases}$
(1)记$b_{n}=a_{2n}$，写出$b_{1}$，$b_{2}$，并求数列$\{ b_{n}\}$的通项公式；
(2)求$\{ a_{n}\}$的前$20$项和。
$a_{n + 1}=\begin{cases}a_{n}+1,n为奇数,\\a_{n}+2,n为偶数.\end{cases}$
(1)记$b_{n}=a_{2n}$，写出$b_{1}$，$b_{2}$，并求数列$\{ b_{n}\}$的通项公式；
(2)求$\{ a_{n}\}$的前$20$项和。
答案：
18.【解】
(1)方法一显然$2n$为偶数,
则$a_{2n + 1}=a_{2n}+2,a_{2n + 2}=a_{2n + 1}+1$,
所以$a_{2n + 2}=a_{2n}+3$,即$b_{n + 1}=b_{n}+3$,且$b_{1}=a_{2}=a_{1}+1 = 2$,
所以$\{ b_{n}\}$是以2为首项,3为公差的等差数列,
于是$b_{1}=2,b_{2}=5,b_{n}=3n - 1$.
方法二由题意,得$a_{1}=1,a_{2}=2,a_{3}=4$,
所以$b_{1}=a_{2}=2,b_{2}=a_{4}=a_{3}+1 = 5$.
由$a_{n + 1}-a_{n}=1(n$为奇数)及$a_{n + 1}-a_{n}=2(n$为偶数)可知,
数列从第一项起,
若$n$为奇数,则其后一项减去该项的差为1,
若$n$为偶数,则其后一项减去该项的差为2,
所以$a_{n + 2}-a_{n}=3$,所以$a_{2n + 2}-a_{2n}=3$,即$b_{n + 1}-b_{n}=3$.
所以$b_{n}=b_{1}+(n - 1)×3=3n - 1$.
方法三由题意知,数列$\{ a_{n}\}$满足$a_{1}=1,a_{n + 1}=a_{n}+\frac{3}{2}+\frac{(-1)^{n}}{2}(n\in\mathbf{N}^*)$,
所以$b_{1}=a_{2}=a_{1}+\frac{3}{2}-\frac{1}{2}=1 + 1 = 2$,
$b_{2}=a_{4}=a_{3}+\frac{3}{2}+\frac{1}{2}=a_{3}+1=a_{2}+\frac{3}{2}-\frac{1}{2}+1=a_{2}+2 + 1 = 5$,
则$b_{n}=a_{2n}=(a_{2n}-a_{2n - 1})+(a_{2n - 1}-a_{2n - 2})+·s+(a_{2}-a_{1})+a_{1}$
$=1 + 2 + 1 + 2+·s+1 + 1=n×1+1+(n - 1)=3n - 1$.
所以$b_{1}=2,b_{2}=5$,数列$\{ b_{n}\}$的通项公式为$b_{n}=3n - 1$.
(2)方法一$S_{20}=a_{1}+a_{2}+a_{3}+·s+a_{20}=(a_{1}+a_{3}+a_{5}+·s+a_{19})+(a_{2}+a_{4}+a_{6}+·s+a_{20})=(b_{1}-1 + b_{2}-1 + b_{3}-1+·s+b_{10}-1)+b_{1}+b_{2}+·s+b_{10}=\frac{(b_{1}+b_{10})×10}{2}-10=300$.
方法二由题意知,数列$\{ a_{n}\}$满足$a_{1}=1,a_{2n}=a_{2n - 1}+1$,
$a_{2n + 1}=a_{2n}+2$,
所以$a_{2n + 1}=a_{2n}+2=a_{2n - 1}+1+2=a_{2n - 1}+3$.
所以数列$\{ a_{n}\}$的奇数项是以1为首项,3为公差的等差数列.
同理,由$a_{2n + 2}=a_{2n + 1}+1=a_{2n}+2+1=a_{2n}+3$知数列$\{ a_{n}\}$的偶数项是以2为首项,3为公差的等差数列.
所以数列$\{ a_{n}\}$的前20项和$S_{20}=(a_{1}+a_{3}+a_{5}+·s+a_{19})+(a_{2}+a_{4}+a_{6}+·s+a_{20})=10×1+\frac{10×9}{2}×3+10×2+\frac{10×9}{2}×3=300$.
方法总结
(1)方法一,由题意讨论$\{ b_{n}\}$的性质为最常规的思路和最优的解法;
方法二,利用递推关系式对奇偶两种情况进行分类讨论,然后利用递推关系式确定数列的性质;
方法三,写出数列$\{ a_{n}\}$的通项公式,然后利用累加法求数列$\{ b_{n}\}$的通项公式,是一种更加灵活的思路.
(2)方法一,由通项公式分奇偶的情况求解前$n$项和是一种常规的方法;
方法二,分组求和是一种常见的数列求和的方法,结合等差数列前$n$项和公式和分组的方法进行求和是一种不错的选择.
(1)方法一显然$2n$为偶数,
则$a_{2n + 1}=a_{2n}+2,a_{2n + 2}=a_{2n + 1}+1$,
所以$a_{2n + 2}=a_{2n}+3$,即$b_{n + 1}=b_{n}+3$,且$b_{1}=a_{2}=a_{1}+1 = 2$,
所以$\{ b_{n}\}$是以2为首项,3为公差的等差数列,
于是$b_{1}=2,b_{2}=5,b_{n}=3n - 1$.
方法二由题意,得$a_{1}=1,a_{2}=2,a_{3}=4$,
所以$b_{1}=a_{2}=2,b_{2}=a_{4}=a_{3}+1 = 5$.
由$a_{n + 1}-a_{n}=1(n$为奇数)及$a_{n + 1}-a_{n}=2(n$为偶数)可知,
数列从第一项起,
若$n$为奇数,则其后一项减去该项的差为1,
若$n$为偶数,则其后一项减去该项的差为2,
所以$a_{n + 2}-a_{n}=3$,所以$a_{2n + 2}-a_{2n}=3$,即$b_{n + 1}-b_{n}=3$.
所以$b_{n}=b_{1}+(n - 1)×3=3n - 1$.
方法三由题意知,数列$\{ a_{n}\}$满足$a_{1}=1,a_{n + 1}=a_{n}+\frac{3}{2}+\frac{(-1)^{n}}{2}(n\in\mathbf{N}^*)$,
所以$b_{1}=a_{2}=a_{1}+\frac{3}{2}-\frac{1}{2}=1 + 1 = 2$,
$b_{2}=a_{4}=a_{3}+\frac{3}{2}+\frac{1}{2}=a_{3}+1=a_{2}+\frac{3}{2}-\frac{1}{2}+1=a_{2}+2 + 1 = 5$,
则$b_{n}=a_{2n}=(a_{2n}-a_{2n - 1})+(a_{2n - 1}-a_{2n - 2})+·s+(a_{2}-a_{1})+a_{1}$
$=1 + 2 + 1 + 2+·s+1 + 1=n×1+1+(n - 1)=3n - 1$.
所以$b_{1}=2,b_{2}=5$,数列$\{ b_{n}\}$的通项公式为$b_{n}=3n - 1$.
(2)方法一$S_{20}=a_{1}+a_{2}+a_{3}+·s+a_{20}=(a_{1}+a_{3}+a_{5}+·s+a_{19})+(a_{2}+a_{4}+a_{6}+·s+a_{20})=(b_{1}-1 + b_{2}-1 + b_{3}-1+·s+b_{10}-1)+b_{1}+b_{2}+·s+b_{10}=\frac{(b_{1}+b_{10})×10}{2}-10=300$.
方法二由题意知,数列$\{ a_{n}\}$满足$a_{1}=1,a_{2n}=a_{2n - 1}+1$,
$a_{2n + 1}=a_{2n}+2$,
所以$a_{2n + 1}=a_{2n}+2=a_{2n - 1}+1+2=a_{2n - 1}+3$.
所以数列$\{ a_{n}\}$的奇数项是以1为首项,3为公差的等差数列.
同理,由$a_{2n + 2}=a_{2n + 1}+1=a_{2n}+2+1=a_{2n}+3$知数列$\{ a_{n}\}$的偶数项是以2为首项,3为公差的等差数列.
所以数列$\{ a_{n}\}$的前20项和$S_{20}=(a_{1}+a_{3}+a_{5}+·s+a_{19})+(a_{2}+a_{4}+a_{6}+·s+a_{20})=10×1+\frac{10×9}{2}×3+10×2+\frac{10×9}{2}×3=300$.
方法总结
(1)方法一,由题意讨论$\{ b_{n}\}$的性质为最常规的思路和最优的解法;
方法二,利用递推关系式对奇偶两种情况进行分类讨论,然后利用递推关系式确定数列的性质;
方法三,写出数列$\{ a_{n}\}$的通项公式,然后利用累加法求数列$\{ b_{n}\}$的通项公式,是一种更加灵活的思路.
(2)方法一,由通项公式分奇偶的情况求解前$n$项和是一种常规的方法;
方法二,分组求和是一种常见的数列求和的方法,结合等差数列前$n$项和公式和分组的方法进行求和是一种不错的选择.
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