1.有若干个数,第一个数记为$a_1$,第二个数记为$a_2$,第三个数记为$a_3$,$···$,第$n$个数记为$a_n$.若$a_1=10$,从第二个数起,每个数都等于1与它前面那个数的差的倒数,则$a_{2024}$的值为 (
A.$-\dfrac{1}{9}$
B.$\dfrac{9}{10}$
C.$10$
D.$\dfrac{10}{9}$
A
)A.$-\dfrac{1}{9}$
B.$\dfrac{9}{10}$
C.$10$
D.$\dfrac{10}{9}$
答案：1.A 点拨:因为$a_1=10$,$a_2=\dfrac{1}{1-10}=-\dfrac{1}{9}$,$a_3=\dfrac{1}{1-(-\dfrac{1}{9})}=\dfrac{9}{10}$,$a_4=\dfrac{1}{1-\dfrac{9}{10}}=10$,$\dots$,所以$a_1,a_2,a_3,a_4,\dots,a_n$以$10$,$-\dfrac{1}{9}$,$\dfrac{9}{10}$为一个循环组无限循环下去.
因为$2024÷3=674······2$,所以$a_{2024}$的值与$a_2$的值相同,即为$-\dfrac{1}{9}$.
因为$2024÷3=674······2$,所以$a_{2024}$的值与$a_2$的值相同,即为$-\dfrac{1}{9}$.
2. 对于 $ x＞0 $，规定 $ f(x)=\frac{x}{x+1} $，例如 $ f(2)=\frac{2}{2+1}=\frac{2}{3} $，$ f(\frac{1}{2})=\frac{\frac{1}{2}}{\frac{1}{2}+1}=\frac{1}{3} $，那么 $ f(\frac{1}{2024})+f(\frac{1}{2023})+\dots +f(\frac{1}{3})+f(\frac{1}{2})+f(1)+f(2)+\dots +f(2024)= $
2023.5
答案：2.2023.5 点拨:因为$f(1)=\dfrac{1}{1+1}=\dfrac{1}{2}$,$f(2)=\dfrac{2}{2+1}=\dfrac{2}{3}$,$f(\dfrac{1}{2})=\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}+1}=\dfrac{1}{3}$,$f(3)=\dfrac{3}{3+1}=\dfrac{3}{4}$,$f(\dfrac{1}{3})=\dfrac{\dfrac{1}{3}}{\dfrac{1}{3}+1}=\dfrac{1}{4}$,$\dots$,所以$f(2)+f(\dfrac{1}{2})=1$,$f(3)+f(\dfrac{1}{3})=1$,$\dots$,所以原式$=f(1)+[f(2)+f(\dfrac{1}{2})]+[f(3)+f(\dfrac{1}{3})]+\dots+[f(2024)+f(\dfrac{1}{2024})]=0.5+1×2023=2023.5$.
3.定义:$a$是不为1的有理数,我们把$\frac{1}{1-a}$称为$a$的差倒数.例如:2的差倒数是$\frac{1}{1-2}=-1$,$-1$的差倒数是$\frac{1}{1-(-1)}=\frac{1}{2}$.已知$a_1=-\frac{1}{3}$,$a_2$是$a_1$的差倒数,$a_3$是$a_2$的差倒数,$a_4$是$a_3$的差倒数,……依此类推,解答下列问题:
(1)$a_2=$
(2)求$a_1+a_2+a_3+\dots+a_{2025}$的值.
(1)$a_2=$
$\dfrac{3}{4}$
,$a_3=$$4$
,$a_4=$$-\dfrac{1}{3}$
;(2)求$a_1+a_2+a_3+\dots+a_{2025}$的值.
答案：3.(1)$\dfrac{3}{4}$ $4$ $-\dfrac{1}{3}$ 点拨:由题意可得,当$a_1=-\dfrac{1}{3}$时,$a_2=\dfrac{1}{1-(-\dfrac{1}{3})}=\dfrac{3}{4}$,$a_3=\dfrac{1}{1-\dfrac{3}{4}}=4$,$a_4=\dfrac{1}{1-4}=-\dfrac{1}{3}$.
(2)解:由(1)知这列数以$-\dfrac{1}{3}$,$\dfrac{3}{4}$,$4$为一个循环组,依次出现.
因为$-\dfrac{1}{3}+\dfrac{3}{4}+4=-\dfrac{4}{12}+\dfrac{9}{12}+\dfrac{48}{12}=\dfrac{53}{12}$,$2025÷3=675$,所以$a_1+a_2+a_3+\dots+a_{2025}=(a_1+a_2+a_3)+\dots+(a_{2023}+a_{2024}+a_{2025})=\dfrac{53}{12}×675=\dfrac{11925}{4}$.
(2)解:由(1)知这列数以$-\dfrac{1}{3}$,$\dfrac{3}{4}$,$4$为一个循环组,依次出现.
因为$-\dfrac{1}{3}+\dfrac{3}{4}+4=-\dfrac{4}{12}+\dfrac{9}{12}+\dfrac{48}{12}=\dfrac{53}{12}$,$2025÷3=675$,所以$a_1+a_2+a_3+\dots+a_{2025}=(a_1+a_2+a_3)+\dots+(a_{2023}+a_{2024}+a_{2025})=\dfrac{53}{12}×675=\dfrac{11925}{4}$.