a) Ta có: \({u_{n + 1}} = 3\left( {n + 1} \right) - 1 = 3n + 2\).

Suy ra \({u_{n + 1}} > {u_n}\).

b) Ta có: \({v_{n + 1}} = \frac{1}{{{{\left( {n + 1} \right)}^2}}}\).

Suy ra: \({u_{n + 1}} < {u_n}\).

uses crt;

var a:array[1..100]of integer;   

i,n,t,t1,t2,t3,min,max:integer;   

kt:boolean;

begin

clrscr;

write('Nhap n='); readln(n);

for i:=1 to n do 

begin     

write('A[',i,']='); readln(a[i]); 

end;

writeln('Day so ban vua nhap la: ');

for i:=1 to n do 

write(a[i]:4);

writeln;

t:=0;

for i:=1 to n do 

t:=t+a[i];

kt:=true;

for i:=2 to trunc(sqrt(t)) do 

if t mod i=0 then     

begin       

kt:=false;       

break;     

end;

writeln('Tong cua day so la: ',t);

if (kt=true) and (t>1) then writeln(t,' la so nguyen to')

else writeln(t,' khong la so nguyen to');

writeln('Cac so o vi tri le cua day so la: ');

t1:=0;

for i:=1 to n do 

if i mod 2=1 then     

begin       

write(a[i]:4);       

t1:=t1+a[i];     

end;

writeln;

writeln('Tong cac so o vi tri le cua day so la: ',t1);

if trunc(sqrt(t1))=sqrt(t1) then writeln(t1,' la so chinh phuong')

else writeln(t1,' khong la so chinh phuong');

writeln('Cac so o vi tri chan cua day so la: ');

t2:=0;

for i:=1 to n do 

if i mod 2=0 then     

begin       

write(a[i]:4);       

t2:=t2+a[i];     

end;

writeln;

writeln('Tong cac so o vi tri chan cua day so la: ',t2);

t3:=0;

for i:=1 to t2 do 

if t2 mod i=0 then t3:=t3+i;

if t3=t2 then writeln(t2,' la so hoan hao')

else writeln(t2,' khong la so hoan hao');

max:=a[1];

min:=a[1];

for i:=1 to n do 

begin     

if max<a[i] then max:=a[i];     

if min>a[i] then min:=a[i]; 

end;

writeln('So lon nhat cua day la: ',max);

writeln('So nho nhat cua day la: ',min);

writeln('Tong cua so lon nhat va so nho nhat la: ',max+min);

readln;

end.

uses crt;

var a:array[1..100]of integer;

n,i,t,kt,j,t1,t2,t3,max,min:integer;

begin

clrscr;

write('Nhap n='); readln(n);

for i:=1 to n do 

begin

write('A[',i,']='); readln(a[i]);

end;

writeln('Day so ban vua nhap la: ');

for i:=1 to n do 

  write(a[i]:4);

writeln;

t:=0;

for i:=1 to n do 

  t:=t+a[i];

writeln('Tong cua day so vua nhap la: ',t);

kt:=0;

for i:=2 to trunc(sqrt(t)) do 

  if t mod i=0 then

begin

kt:=1;

break;

end;

if (kt=0) and (t>1) then writeln(t,' la so nguyen to');

else writeln(t,' khong la so nguyen to');

t1:=0;

writeln('Cac so o vi tri le cua day la: ');

for i:=1 to n do 

  if i mod 2=1 then 

begin

write(a[i]:4);

t1:=t1+a[i];

end;

writeln;

writeln('Tong cac so o vi tri le la: ',t1);

if trunc(sqrt(t1))=sqrt(t1) then writeln(t1,' la so chinh phuong')

else writeln(t1,' khong la so chinh phuong');

t2:=0;

writeln('Cac so o vi tri chan la: ');

for i:=1 to n do 

  if i mod 2=0 then 

begin

write(a[i]:4);

t2:=t2+a[i];

end;

writeln;

writeln('Tong cac so o vi tri chan la: ',t2);

t3:=0;

for i:=1 to t2 do 

if t2 mod i=0 then t3:=t3+i;

if t3=t2 then writeln(t2,' la so hoan hao')

else writeln(t2,' khong la so hoan hao');

max:=a[1];

min:=a[1];

for i:=1 to n do 

begin

if max<a[i] then max:=a[i];

if min>a[i] then min:=a[i];

end;

writeln('Gia tri lon nhat la: ',max);

writeln('Gia tri nho nhat la: ',min);

writeln('Tong cua hai so lon nhat va nho nhat la: ',max+min);

readln;

end.

Chọn A.

Ta có: 

Do đó: 

- Ta chứng minh dãy (y_n)  tăng.

Ta có: 

- Ta chứng minh dãy (y_n)  bị chặn.

Trước hết ta chứng minh: x_n ≤ 4(n – 1) (1)

 * Với n = 2, ta có: x₂ = 4x₁ = 4 nên (1) đúng với n = 2

 * Giả sử (1) đúng với n, tức là: x_n ≤ 4(n – 1), ta có

Nên (1) đúng với n + 1. Theo nguyên lí quy nạp ta suy ra (1) đúng

Ta có: 

Vậy bài toán được chứng minh.

Chọn D

Bạn xem lại xem viết đề có thiếu/nhầm gì không?

\(u_n=\dfrac{n+2}{n}\)

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

\(\Rightarrow u_{n+1}-u_n=\dfrac{n+3}{n+1}-\dfrac{n+2}{n}\)

\(\Rightarrow u_{n+1}-u_n=\dfrac{n\left(n+3\right)-\left(n+1\right)\left(n+2\right)}{n\left(n+1\right)}\)

\(\Rightarrow u_{n+1}-u_n=\dfrac{n^2+3n-\left(n^2+3n+2\right)}{n\left(n+1\right)}\)

\(\Rightarrow u_{n+1}-u_n=\dfrac{n^2+3n-n^2-3n-2}{n\left(n+1\right)}\)

\(\Rightarrow u_{n+1}-u_n=\dfrac{-2}{n\left(n+1\right)}< 0\)

Vậy dãy số \(u_n\) đã cho là dãy giảm

\(u_n=\sqrt[]{n+10}-\sqrt[]{n+2}\)

\(\Leftrightarrow u_n=\dfrac{n+10-\left(n+2\right)}{\sqrt[]{n+10}+\sqrt[]{n+2}}\)

\(\Leftrightarrow u_n=\dfrac{8}{\sqrt[]{n+10}+\sqrt[]{n+2}}\)

\(u_{n+1}=\sqrt[]{n+11}-\sqrt[]{n+3}\)

\(\Leftrightarrow u_{n+1}=\dfrac{n+11-\left(n+3\right)}{\sqrt[]{n+11}+\sqrt[]{n+3}}\)

\(\Leftrightarrow u_{n+1}=\dfrac{8}{\sqrt[]{n+11}+\sqrt[]{n+3}}\)

\(u_{n+1}-u_n=8\left(\dfrac{1}{\sqrt[]{n+11}+\sqrt[]{n+3}}-\dfrac{1}{\sqrt[]{n+10}+\sqrt[]{n+2}}\right)\)

mà \(\dfrac{1}{\sqrt[]{n+11}+\sqrt[]{n+3}}< \dfrac{1}{\sqrt[]{n+10}+\sqrt[]{n+2}}\)

\(\Rightarrow u_{n+1}-u_n< 0\)

Vậy dãy đã cho là dãy số giảm

\(u_n=\dfrac{3^n-1}{2^n}\)

\(\Rightarrow u_{n+1}=\dfrac{3^{n+1}-1}{2^{n+1}}\)

\(\Rightarrow u_{n+1}-u_n=\dfrac{3^{n+1}-1}{2^{n+1}}-\dfrac{3^n-1}{2^n}\)

\(\Rightarrow u_{n+1}-u_n=\dfrac{2^n.3^{n+1}-2^n-2^{n+1}.3^n+2^{n+1}}{2^n.2^{n+1}}\)

\(=\dfrac{2^n.3^n\left(3-2\right)-2^n\left(2-1\right)}{2^{2n+1}}\)

\(=\dfrac{2^n.\left(3^n-1\right)}{2^{2n+1}}\)

\(=\dfrac{\left(3^n-1\right)}{2}>0\left(n>1\right)\)

Vậy dãy \(u_n\)đã cho tăng