数列几题

$例1、已知数列\{a_n\}中，a_1=2,a_{n+1}=a_n+n+1.$
$(1)求数列\{a_n\}的通项公式；$
$(2)设b_n=\cfrac{1}{a_n},数列\{b_n\}的前n项和为T_n，证明T_n\lt 2$

$(1)a_{n+1}=a_n+n+1，有递推关系式：$
$a_{n}=a_{n-1}+(n-1)+1$
$a_{n-1}=a_{n-2}+(n-2)+1$
$a_{n-2}=a_{n-3}+(n-3)+1$
$a_{n-3}=a_{n-4}+(n-4)+1$
$\cdots \cdots \cdots\cdots$
$a_3=a_2+2+1$
$a_2=a_1+1+1$
累加之
$a_{n}+a_{n-1}+a_{n-2}+a_{n-3}+\cdots +a_3+a_2=a_{n-1}+a_{n-2}+a_{n-3}+a_{n-4}+\cdots +a_2+a_1$
$+(n-1)+(n-2)+(n-3)+(n-4)+\cdots +1+1+1\cdots +1$
$a_n=a_1+(n-1)+(n-2)+(n-3)+(n-4)+\cdots+2+1(n-1)\times 1$
$=2+(n-1)+(n-2)+(n-3)+(n-4)+\cdots+2+1+(n-1)$
$=n+(n-1)+\cdots +1+1=\cfrac{n(n+1)}{2}+1=\cfrac{n^2+n+2}{2}$
$b_n=\cfrac{1}{a_n}=\cfrac{2}{n^2+n+2}\lt {\color{Red} \cfrac{2}{n(n+1)}=2(\cfrac{1}{n}-\cfrac{1}{n+1})} $
$b_1\lt 2(1-\cfrac{1}{2})$
$b_2\lt 2(\cfrac{1}{2}-\cfrac{1}{3})$
$b_3\lt 2(\cfrac{1}{3}-\cfrac{1}{4})$
$\cdots\cdots\cdots\cdots$
$b_{n-2}\lt 2(\cfrac{1}{n-2}-\cfrac{1}{n-1})$
$b_{n-1}\lt 2(\cfrac{1}{n-1}-\cfrac{1}{n})$
$b_n\lt 2(\cfrac{1}{n}-\cfrac{1}{n+1})$
$T_n\lt 2(1-\cfrac{1}{n+1})\lt 2$

$例2、已知数列\{a_n\}的前n项和为S_n,且a_2=3,a_{n+1}=S_n+n+1$
$(1)证明数列\{a_n+1\}是等比数列$
$递推公式：a_{n+1}=S_n+n+1\quad ①\Rightarrow a_2=S_1+1+1=a_1+2,a_1=1$
$\qquad\qquad a_{n}=S_{n-1}+(n-1)+1\quad ②$
$①-②,得a_{n+1}－a_{n}＝S_n+n+1－[S_{n-1}+(n-1)+1]$
$a_{n+1}－a_{n}=a_n+1\Rightarrow a_{n+1}+1=2(a_n+1)$
$故数列\{a_n+1\}是以a_1+1=2为首项，q=2的等比数列$

$15、d=\cfrac{S_2}{2}-\cfrac{S_1}{1}=\cfrac{a_2+a_1}{2}-\cfrac{a_1}{1}=\cfrac{3-a_1}{2}$
$\cfrac{S_5}{5}=S_1+4d=a_1+4d=a_1+4\times \cfrac{3-a_1}{2}=6-a_1=5\Rightarrow a_1=1,d=1$
$\cfrac{S_n}{n}=n\Rightarrow S_n=n^2\Rightarrow n\ge 2,S_{n-1}=(n-1)^2,S_n-S_{n-1}=a_n=2n-1(n\ge 2)$
$n=1时，a_1=1,故a_n=2n-1在n=1时也成立。$
$(2)、b_n=2(n+1)(a_n+1)=4n(n+1)\Rightarrow \cfrac{1}{b_n}=\cfrac{1}{4n(n+1)}=\cfrac{1}{4}(\cfrac{1}{n}-\cfrac{1}{n+1})$
$\cfrac{1}{b_1}+\cfrac{1}{b_2} +\cdots +\cfrac{1}{b_n}=\cfrac{1}{4}(1-\cfrac{1}{2}+\cfrac{1}{2}-\cfrac{1}{3}+\cfrac{1}{3}-\cfrac{1}{4}+\cdots+\cfrac{1}{n-1}-\cfrac{1}{n}+\cfrac{1}{n}-\cfrac{1}{n+1})$
$=\cfrac{1}{4}(1-\cfrac{1}{n+1})\lt \cfrac{1}{4}$

26年江苏省高三G4联考第17题
$已知数列\{a_n\}满足a_1=1,\cfrac{S_{n+1}}{S_n}=\cfrac{n+2}{n}$
$(1)求数列\{a_n\}的通项公式；$
$(2)设b_n=a_n^2(\cos^2\cfrac{n\pi}{3}-\sin^2\cfrac{n\pi}{3}),求数列\{b_n\}的前n项和T_n$
$解:\cfrac{S_{n+1}}{S_n}=\cfrac{n+2}{n}\Rightarrow \cfrac{S_{n+1}}{S_n}-1=\cfrac{n+2}{n}-1\Rightarrow \cfrac{a_{n+1}}{S_n}=\cfrac{2}{n}\Rightarrow na_{n+1}=2S_n\quad {\color{Red} ①}$
$由递推公式得到(n-1)a_n=2S_{n-1}\quad {\color{Red} ②} \Rightarrow {\color{Red}①-② } ,得na_{n+1}-(n-1)a_n=2a_n$
$na_{n+1}=(n+1)a_n\Rightarrow \cfrac{a_{n+1}}{n+1}=\cfrac{a_{n}}{n}=\cfrac{a_1}{1}=1\Rightarrow a_n=n$
${\color{Red}方法2}:S_1\times \cfrac{S_2}{S_1}\times \cfrac{S_3}{S_2}\times \cfrac{S_4}{S_3}\times \cfrac{S_5}{S_4}\times\cdots\times \cfrac{S_{n+1}}{S_n}=1\times\cfrac{3}{1}\times\cfrac{4}{2} \times\cfrac{5}{3} \times\cfrac{6}{4} \times\cfrac{7}{5}\times \cdots\times \cfrac{n}{n-2}\times\cfrac{n+1}{n-1}\times\cfrac{n+2}{n} $
$S_{n+1}=\cfrac{(n+1)(n+2)}{2}, S_{n}=\cfrac{n(n+1)}{2},n\ge 2时，a_n=S_{n}-S_{n-1}=n,显然n=1也成立。$

$b_n=a_n^2(\cos^2\cfrac{n\pi}{3}-\sin^2\cfrac{n\pi}{3})=n^2\cos \cfrac{2n\pi}{3}$
$b_1=\cos \cfrac{2\pi}{3}=1^2\times(-\cfrac{1}{2})$
$b_2=2^2\times\cos \cfrac{4\pi}{3}=2^2\times(-\cfrac{1}{2})$
$b_3=3^3\times\cos \cfrac{6\pi}{3}=3^2\times(+1)$
$b_4=4^2\times\cos \cfrac{8\pi}{3}=4^2\times(-\cfrac{1}{2})$
$\cdots\cdots\cdots\cdots\cdots$
$①设n=3k时，T_n=-{\color{Red}\cfrac{1}{2}}\times\cfrac{3k(3k+1)(6k+1)}{6} +{\color{Red}\cfrac{3}{2}}\times 9\times \cfrac{k(k+1)(2k+1)}{6}=$
$=-\cfrac{1}{4}k(3k+1)(6k+1)+\cfrac{9}{4}k(k+1)(2k+1)=k[-\cfrac{1}{4}(3k+1)(6k+1)+\cfrac{9}{4}(k+1)(2k+1)]$
$=k[-\cfrac{18}{4}k^2-\cfrac{9}{4}k-\cfrac{1}{4}+\cfrac{18}{4}k^2+\cfrac{9}{4}\times 3k+\cfrac{9}{4}]$
$=k(\cfrac{9}{2}k+2)=\cfrac{9}{2}k^2+2k=\cfrac{9}{2}(\cfrac{n}{3})^2+\cfrac{2n}{3}=\cfrac{3n^2+4n}{6}$
$设②n=3k+1,T_n=\cfrac{9}{2}k^2+2k-{\color{Red}\cfrac{1}{2}(3k+1)^2}=\cfrac{9}{2}k^2+2k-\cfrac{1}{2}(9k^2+6k+1)$
$=-k-\cfrac{1}{2}\quad k=\cfrac{n-1}{3}\Rightarrow T_n=-\cfrac{n-1}{3}-\cfrac{1}{2}=-\cfrac{2n+1}{6}$
$③设n=3k+2,T_n=-k-\cfrac{1}{2}-\cfrac{1}{2}(3k+2)^2=-k-\cfrac{1}{2}-\cfrac{1}{2}\times n^2=-\cfrac{3n^2+2n+1}{6}$
${\color{Green}这里是错的,k值不同于 ②} $
$k=\cfrac{n-2}{3}\quad T_n=-k-\cfrac{1}{2}-\cfrac{1}{2}(3k+2)^2=-\cfrac{n-2}{3}-\cfrac{1}{2}-\cfrac{1}{2}\times n^2=-\cfrac{3n^2+2n-1}{6}$
$综合上述T_n=\begin{cases} \cfrac{3n^2+4n}{6}\qquad\quad n=3k时\\ -\cfrac{2n+1}{6}\qquad\quad n=3k+1时\\-\cfrac{3n^2+2n-1}{6}\quad n=3k+2\end{cases}$

${\color{Red}利用完全立方差或立方差公式推导a_n=n^2的前n项和S_n } $
$(n-1)^3＝n^2-3n^2+3n-1\Rightarrow n^3-(n-1)^3＝3n^2-3n+1$
$n^3-(n-1)^3＝3n^2-3n+1$
$(n-1)^3-(n-2)^3＝3(n-1)^2-3n(n-1)+1$
$(n-2)^3-(n-3)^3＝3(n-2)^2-3n(n-2)+1$
$(n-3)^3-(n-4)^3＝3(n-3)^2-3n(n-3)+1$
$\cdots \cdots \cdots$
$2^3-1^3＝3\cdot 2^2-3\times 2+1$
$1^3-0^3＝3\cdot 1^2-3\times 1+1$
左右两边相加
$n^3=3S_n-3\times \cfrac{n(n+1)}{2}+n$
$3S_n=n^3+3\times \cfrac{n(n+1)}{2}-n=n[n^2+\cfrac{3}{2}(n+1)-1]$
$\cfrac{n}{2}(2n^2+3n+1)=\cfrac{n}{2}(2n+1)(n+1)$
$S_n=\cfrac{1}{6}\times n(n+1)(2n+1)$

数列讨论奇偶求通项－构造相邻项

$21年新高考一卷，已知数列\{a_n\}满足a_1=1,a_{n+1}=\begin{cases}a_n+1,n为奇数\\a_n+2,n为偶数\end{cases}$
$(1)记b_n=a_{2n},写出b_1,b_2,并求数列\{b_n\}的通项公式；$
$(2)求\{a_n\}的前20项和$
$a_2=a_1+1=1+1=2,b_1=a_2=2,a_3=a_2+2=4,b_2=a_4=a_3+1=5$
${\color{Green} \because \quad b_n=a_{2n}}$
$\therefore b_{n+1}=a_ {2(n+1)}=a_ {{\color{Green} 2n+1} +1}=a_ {{\color{Red} 2n}+1}+1=a_{{\color{Red} 2n} }+2+1=b_n+3$
$\Rightarrow b_{n+1}=b_n+3,故b_n为首项为2,公差为3的等差数列,b_n=2+3(n-1)=3n-1$
$(2)求前20项和，需分别计算奇数项和偶数项和；$
$偶数项：共10项(a_2,a_4,a_6\cdots a_{20}),b_{10}=3\times 10-1=29,$
$S_偶=\cfrac{10(b_1+b_{10})}{2}=\cfrac{10(2+29)}{2}=155$
$奇数项：共10项(a_1,a_3,a_5\cdots a_{19})$
$设\{c_k\},(c_k=a_{2k-1},k\in 1,2,3,\cdots 10)$
$c_{k+1}=a_{2k+1}=a_{2k}+2=(a_{2k-1}+1)+2=a_{2k-1}+3=c_k+3$
$\{c_k\}是首项为c_1=1,公差d=3的等差数列，和为S_奇=\cfrac{10(C_1+C_{10})}{2}=145$

$例6.记S_n为正项数列\{a_n\}的前n项和，且a_1=\cfrac{1}{3},2S_n=(3^n-1)a_n;$
$(1)求数列\{a_n\}的通项公式;$
$(2)设b_n=\cfrac{a_{n+1}}{(1-a_n)S_{n+1}},记数列\{b_n\}的前n项和为T_n,证明:T_n\ge \cfrac{3}{8}$
$解:2S_n=(3^n-1)a_n\quad ① n\ge 2时，2S_{n-1}=(3^{n-1}-1)a_{n-1}\quad ②$
$①-②,得2a_n=(3^n-1)a_n-(3^{n-1}-1)a_{n-1}\Rightarrow (3^{n-1}-1)a_{n-1}=(3^n-3)a_n$
$\Rightarrow \cfrac{a_n}{a_{n-1}}=\cfrac{3^{n-1}-1}{3^n-3}=\cfrac{1}{3},所以a_n为首项\cfrac{1}{3}，公式q为\cfrac{1}{3}的等比数列$
$a_n=(\cfrac{1}{3})^n,S_n=\cfrac{\cfrac{1}{3}-(\cfrac{1}{3})^{n+1}}{1-\cfrac{1}{3}}=\cfrac{1-(\cfrac{1}{3})^n}{2}$
$b_n=\cfrac{a_{n+1}}{(1-a_n)S_{n+1}}=\cfrac{(\cfrac{1}{3})^{n+1}}{[1-(\cfrac{1}{3})^{n}]\cfrac{1-(\cfrac{1}{3})^{n+1}}{2}}=\cfrac{2\times (\cfrac{1}{3})^{n+1}}{[1-(\cfrac{1}{3})^{n}][1-(\cfrac{1}{3})^{n+1}]}$
${\color{Orange} \because \quad \cfrac{1}{1-(\cfrac{1}{3})^{n}}-\cfrac{1}{1-(\cfrac{1}{3})^{n+1}}} =\cfrac{1-(\cfrac{1}{3})^{n+1}-[{1-(\cfrac{1}{3})^{n}}]}{[1-(\cfrac{1}{3})^{n}][1-(\cfrac{1}{3})^{n+1}]}$
$=\cfrac{-(\cfrac{1}{3})^{n+1}{+(\cfrac{1}{3})^{n}}}{[1-(\cfrac{1}{3})^{n}][1-(\cfrac{1}{3})^{n+1}]}=\cfrac{{\cfrac{2}{3}\times (\cfrac{1}{3})^{n}}}{[1-(\cfrac{1}{3})^{n}][1-(\cfrac{1}{3})^{n+1}]}=\cfrac{{2\times (\cfrac{1}{3})^{n+1}}}{[1-(\cfrac{1}{3})^{n}][1-(\cfrac{1}{3})^{n+1}]}$
${\color{Orange} \therefore \quad b_n=\cfrac{1}{1-(\cfrac{1}{3})^{n}}-\cfrac{1}{1-(\cfrac{1}{3})^{n+1}}} $
$T_n=b_1+b_2+b_3+\cdots +b_n={\color{Orange} \cfrac{1}{1-\cfrac{1}{3}}-\cfrac{1}{1-(\cfrac{1}{3})^2}}+\cfrac{1}{1-(\cfrac{1}{3})^{2}}-\cfrac{1}{1-(\cfrac{1}{3})^{3}}$
$+{\color{Orange} \cfrac{1}{1-(\cfrac{1}{3})^{3}}-\cfrac{1}{1-(\cfrac{1}{3})^{4}}} +\cfrac{1}{1-(\cfrac{1}{3})^{4}}-\cfrac{1}{1-(\cfrac{1}{3})^{5}}+\cdots +\cfrac{1}{1-(\cfrac{1}{3})^{n}}-\cfrac{1}{1-(\cfrac{1}{3})^{n+1}}$
$=\cfrac{1}{1-\cfrac{1}{3}}-\cfrac{1}{1-(\cfrac{1}{3})^{n+1}},要证\cfrac{3}{2}-\cfrac{1}{1-(\cfrac{1}{3})^{n+1}}\ge \cfrac{3}{8}即证\cfrac{1}{1-(\cfrac{1}{3})^{n+1}}\le \cfrac{9}{8}$
$即证1-(\cfrac{1}{3})^{n+1}\ge \cfrac{8}{9}=1-\cfrac{1}{9},n=1即可$
$T_n=\cfrac{1}{1-\cfrac{1}{3}}-\cfrac{1}{1-(\cfrac{1}{3})^{n+1}}=\cfrac{3}{2}-\cfrac{1}{1-(\cfrac{1}{3})^{n+1}},T_n\nearrow,T_n\ge (T_n)_{min}=T_1=\cfrac{3}{2}-\cfrac{9}{8}=\cfrac{3}{8}$

$例7、数列\{a_n\}的前n项和为S_n,且a_1=１,\{\cfrac{S_n}{a_n}\}是公差为\cfrac{1}{3}的等差数列$
$(1)求\{a_n\}通项公式$
$(2)证明:\cfrac{1}{a_1}+\cfrac{1}{a_2}+\cdots+\cfrac{1}{a_n}\lt 2$
$解:\cfrac{S_n}{a_n}=1+\cfrac{1}{3}(n-1)=\cfrac{n+2}{3}\Rightarrow S_n=\cfrac{n+2}{3}a_n\quad ①$
$n\ge 2时，有S_{n-1}=\cfrac{n-1+2}{3}a_{n-1}=\cfrac{n+1}{3}a_{n-1}\quad②$
$①-②,得\cfrac{a_n}{a_{n-1}}=\cfrac{n+1}{n-1}$
$\cfrac{a_2}{a_1}=\cfrac{2+1}{2-1}=\cfrac{3}{1}$
$\cfrac{a_3}{a_2}=\cfrac{3+1}{3-1}=\cfrac{4}{2}$
$\cfrac{a_4}{a_3}=\cfrac{4+1}{4-1}=\cfrac{5}{3}$
$\cdots \cdots\cdots\cdots $
$\cfrac{a_{n-1}}{a_{n-2}}=\cfrac{n-1+1}{n-1-1}=\cfrac{n}{n-2}$
$\cfrac{a_n}{a_{n-1}}=\cfrac{n+1}{n-1}$
$\cfrac{a_n}{a_{n-1}}\times\cfrac{a_{n-1}}{a_{n-2}}\cdots \cfrac{a_4}{a_3}\times \cfrac{a_3}{a_2}\times \cfrac{a_2}{a_1}=a_n=\cfrac{n+1}{n-1}\times\cfrac{n}{n-2}\cdots\cfrac{5}{3}\times \cfrac{4}{2}\times \cfrac{3}{1}=\cfrac{n(n+1)}{2}$
${\color{Red} \because \quad} \cfrac{1}{a_n}=\cfrac{2}{n(n+1)}=2(\cfrac{1}{n}-\cfrac{1}{n+1})$
${\color{Red} \therefore\quad} \cfrac{1}{a_1}+\cfrac{1}{a_2}+\cdots+\cfrac{1}{a_n}=2(1-\cfrac{1}{2}+\cfrac{1}{2}-\cfrac{1}{3}+\cdots +\cfrac{1}{n}-\cfrac{1}{n+1})=2(1-\cfrac{1}{n+1})\lt 2$

$例8、已知正项数列\{a_n\}的前n项和S_n，满足S_n=(\cfrac{a_n+1}{2})^2.$
$(1)求数列\{a_n\}的通项公式;$
$(2)记b_n=\cfrac{n+1}{S_nS_{n+2}},设数列\{b_n\}的前n项和为T_n.求证T_n\lt \cfrac{5}{16}$
$解：(1).a_1=S_1=(\cfrac{a_1+1}{2})^2\Rightarrow a_1=1$
$n\ge 2时，S_{n-1}=(\cfrac{a_{n-1}+1}{2})^2$
$a_n=S_n-S_{n-1}=(\cfrac{a_n+1}{2})^2-(\cfrac{a_{n-1}+1}{2})^2=\cfrac{a_n+a_{n-1}+2}{2} \times\cfrac{a_n-a_{n-1}+2}{2}$
$\Rightarrow 4a_n=a_n^2-a_{n-1}^2+2(a_n-a_{n-1})\Rightarrow a_n^2-a_{n-1}^2-2(a_n+a_{n-1})=0$
$\Rightarrow (a_n+a_{n-1})(a_n-a_{n-1}-2)=0,a_n+a_{n-1}=0(舍去),a_n-a_{n-1}=2$
$a_{n-1}-a_{n-2}=2$
$a_{n-2}-a_{n-3}=2$
$a_2-a_1=2$
$a_n-a_1=2(n-1)\Rightarrow a_n=2n-1\quad(n\ge 2),a=1$
$(2){\color{Red} \because \quad } S_n=(\cfrac{a_n+1}{2})^2=n^2,所以S_{n+2}=(n+2)^2$
$b_n=\cfrac{n+1}{S_nS_{n+2}}=\cfrac{n+1}{n^2(n+2)^2}$
${\color{Red} \because \quad } \cfrac{1}{n^2(n+2)^2} =\cfrac{1}{4(n+1)}[\cfrac{1}{n^2}-\cfrac{1}{(n+2)^2} ]$
${\color{Red} \therefore \quad } b_n=\cfrac{n+1}{n^2(n+2)^2} =\cfrac{1}{4}[\cfrac{1}{n^2}-\cfrac{1}{(n+2)^2} ]$
$T_n=\cfrac{1}{4}(1-\cfrac{1}{3^2}+\cfrac{1}{2^2}-\cfrac{1}{4^2}+\cfrac{1}{3^2}-\cfrac{1}{5^2}+\cdots+\cfrac{1}{n^2}-\cfrac{1}{(n+2)^2}$
$=\cfrac{1}{4}[1+\cfrac{1}{2^2}-\cfrac{1}{(n+1)^2}-\cfrac{1}{(n+2)^2}]\lt \cfrac{5}{16}$

https://uu.890222.xyz/usr/uploads/2025/12/705270186.[5]: https://uu.890222.xyz/usr/uploads/2025/12/2791294339.png