2:

a: \(u_1=\dfrac{2-1}{1+1}=\dfrac{1}{2}\)

\(u_2=\dfrac{2\cdot2-1}{2+1}=1\)

\(u_3=\dfrac{2\cdot3-1}{3+1}=\dfrac{5}{4}\)

\(u_4=\dfrac{2\cdot4-1}{4+1}=\dfrac{7}{5}\)

b: Đặt \(\dfrac{2n-1}{n+1}=\dfrac{13}{7}\)

=>7(2n-1)=13(n+1)

=>14n-7=13n+13

=>n=20

=>13/7 là số hạng thứ 20 trong dãy

1:

a: u1=1^2-1=0

u2=2^2-1=3

u3=3^2-1=8

u4=4^2-1=15

b: 99=n^2-1

=>n^2=100

mà n>=0

nên n=10

=>99 là số thứ 10 trong dãy

1:

a:

u1=1^2+1=2

u2=2^2+1=5

u3=3^2+1=10

u4=4^2+1=17

b: Đặt 101=n^2+1

=>n^2=100

=>n=10

=>101 là số hạng thứ 10

2:

a: \(u1=\dfrac{1+1}{2-1}=2\)

\(u2=\dfrac{2+1}{2\cdot2-1}=\dfrac{3}{3}=1\)

\(u_3=\dfrac{3+1}{2\cdot3-1}=\dfrac{4}{5}\)

\(u_4=\dfrac{4+1}{2\cdot4-1}=\dfrac{5}{7}\)

b: Đặt \(\dfrac{n+1}{2n-1}=\dfrac{31}{59}\)

=>59(n+1)=31(2n-1)

=>62n-31=59n+59

=>3n=90

=>n=30

=>31/59 là số hạng thứ 30 trong dãy

\(\dfrac{u_{n+1}}{n+1}=3.\dfrac{u_n}{n}\)

Đặt \(\dfrac{u_n}{n}=v_n\Rightarrow\left\{{}\begin{matrix}v_1=\dfrac{1}{3}\\v_{n+1}=3v_n\end{matrix}\right.\)

\(\Rightarrow v_n=\dfrac{1}{3}.3^{n-1}=3^{n-2}\)

\(\Rightarrow S=3^{-1}+3^0+...+3^8=...\)

Đề không cho sẵn dãy tăng à? Vậy phải chứng minh nó tăng trước

\(u_{n+1}=\dfrac{u_n^2+2018u_n+1}{2020}\)

\(u_{n+1}-u_n=\dfrac{u_n^2+2018u_n+1}{2020}-u_n=\dfrac{\left(u_n-1\right)^2}{2020}\ge0\) \(\Rightarrow\) dãy tăng và không bị chặn trên \(\Rightarrow lim\left(u_n\right)=+\infty\)

\(\Rightarrow2020u_{n+1}=u_n^2+2018u_n+1\)

\(\Leftrightarrow2020u_{n+1}-2020=u_n^2+2018u_n-2019\)

\(\Leftrightarrow2020\left(u_{n+1}-1\right)=\left(u_n+2019\right)\left(u_n-1\right)\)

\(\Rightarrow\dfrac{1}{2020\left(u_{n+1}-1\right)}=\dfrac{1}{\left(u_n+2019\right)\left(u_n-1\right)}=\dfrac{1}{2020}\left(\dfrac{1}{u_n-1}-\dfrac{1}{u_n+2019}\right)\)

\(\Rightarrow\dfrac{1}{u_n+2019}=\dfrac{1}{u_n-1}-\dfrac{1}{u_{n+1}-1}\)

Thế n=1;2;...;n ta được:

\(\dfrac{1}{u_1+2019}=\dfrac{1}{u_1-1}-\dfrac{1}{u_2-1}\)

\(\dfrac{1}{u_2+2019}=\dfrac{1}{u_2-1}-\dfrac{1}{u_3-1}\)

...

\(\dfrac{1}{u_n+2019}=\dfrac{1}{u_n-1}-\dfrac{1}{u_{n+1}-1}\)

Cộng vế: \(S_n=\dfrac{1}{u_n-1}-\dfrac{1}{u_{n+1}-1}=\dfrac{1}{2018}-\dfrac{1}{u_{n+1}-1}\)

\(\Rightarrow\lim\left(S_n\right)=\dfrac{1}{2018}-\dfrac{1}{\infty}=\dfrac{1}{2018}\)

\(U_n=\dfrac{\left(n^2-1\right)}{n\left(n+2\right)}U_{n-1}\Rightarrow n\left(n+2\right).U_n=\left(n-1\right)\left(n+1\right).U_{n-1}\)

Đặt \(n\left(n+2\right).U_n=V_n\Rightarrow V_{n-1}=\left(n-1\right)\left(n+2-1\right).U_{n-1}=\left(n-1\right).\left(n+1\right)U_{n-1}\)

\(\Rightarrow V_n=V_{n-1}\)

\(\Rightarrow V_n=V_{n-1}=V_{n-2}=...=V_1\)

Có \(V_1=1.\left(1+2\right).U_1=1\)

\(\Rightarrow V_n=1\)

\(\Rightarrow U_n=\dfrac{V_n}{n\left(n+2\right)}=\dfrac{1}{n\left(n+2\right)}\)

\(\Rightarrow A=\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+...+\dfrac{1}{2015.2017}\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\right)\)

\(=\dfrac{1}{2}\left(1+\dfrac{1}{2}-\dfrac{1}{2016}-\dfrac{1}{2017}\right)\)

\(=...\)

\(u_2=\sqrt{2}\left(2+3\right)-3=5\sqrt{2}-3\)

\(u_3=\sqrt{\dfrac{3}{2}}.5\sqrt{2}-3=5\sqrt{3}-3\)

\(u_4=\sqrt{\dfrac{4}{3}}.5\sqrt{3}-3=5\sqrt{4}-3\)

....

\(\Rightarrow u_n=5\sqrt{n}-3\)

\(\Rightarrow\lim\limits\dfrac{u_n}{\sqrt{n}}=\lim\limits\dfrac{5\sqrt{n}-3}{\sqrt{n}}=5\)