Thuật toán: 

Bước 1: Nhập n

Bước 2: i←1; a←0;

Bước 3: a←a+1/(i*(i+2));

Bước 4: i←i+1;

Bước 5: Nếu i<=n thì quay lại bước 3

Bước 6: xuất a

Bước 7: Kết thúc

Viết chương trình:

uses crt;

var a:real;

i,n:longint;

begin

clrscr;

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

a:=0;

for i:=1 to n do

a:=a+1/(i*(i+2));

writeln(a:4:2);

readln;

end.

Em cảm ơn anh !

Program HOC24;

var b: real;

i,n: integer;

begin

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

b:=0;

for i:=1 to n do b:=b+1/(i+2);

write('B= ',b:1:2);

readln

end.

Sửa lại đề bài : \(A=....+\dfrac{1}{n\left(n+2\right)}\)

Program HOC24;

var i,n: integer;

a: real;

begin

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

a:=0;

for i:=1 to n do a:=a+1/(n*(n+2));

write('A = ',a:6:2);

readln

end.

uses crt;

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

i,n,d:integer;

begin

clrscr;

repeat

writeln('nhap n=');readln(n);

until n>0;

for i:=1 to n do

begin writeln('b[',i,']','=');readln(b[i]);end;

writeln('so cac so le la');

for i:=1 to n do

if b[i] mod 2<>0 then d:=d+1;

writeln(d);readln;end.

uses crt;

var n,i:longint; s:real;

begin

clrscr;

s:=0;

writeln('nhap vao n=');readln(n);

writeln('tong cua A la');

for i:=1 to n do

s:=s + 1/(i*(i+2));

writeln(s:4:3);readln;end.

\(\left(1+\dfrac{1}{1.3}\right).\left(1+\dfrac{1}{2.4}\right).\left(1+\dfrac{1}{3.5}\right).........\left[1+\dfrac{1}{x.\left(x+2\right)}\right]=\dfrac{31}{16}\)

\(\Rightarrow\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}........\dfrac{\left(x+1\right)^2}{x.\left(x+2\right)}=\dfrac{31}{16}\)

\(\Rightarrow\dfrac{\left[2.3.4.............\left(x+1\right)\right].\left[2.3.4.............\left(x+1\right)\right]}{\left(1.2.3...................x\right).\left(3.4.5..........................\left(x+2\right)\right)}=\dfrac{31}{16}\)

\(\Rightarrow\dfrac{\left(x+1\right).2}{1.\left(x+2\right)}=\dfrac{31}{16}\)

\(\Leftrightarrow16.2\left(x+1\right)=31.\left(x+2\right)\)

\(\Rightarrow32x+32=31x+62\)

\(\Rightarrow x=30\)

Vậy x=30

Chúc bn học tốt

thank

\(A=\dfrac{1}{2}\left(2.\dfrac{2}{3}\right)\left(\dfrac{3}{2}.\dfrac{3}{4}\right)\left(\dfrac{4}{3}.\dfrac{4}{5}\right)....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)

\(=\dfrac{2016}{2017}\)

\(A=\dfrac{1}{2}.\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)....\left(\dfrac{1}{2015.2017}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1}.\dfrac{2}{3}\right).\left(\dfrac{3}{2}.\dfrac{3}{4}\right).\left(\dfrac{4}{3}.\dfrac{4}{5}\right)....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)

\(=\dfrac{1}{2}.\left(\dfrac{2}{1}.\dfrac{2}{3}\right).\left(\dfrac{3}{2}.\dfrac{3}{4}\right).\left(\dfrac{4}{3}.\dfrac{4}{5}\right).....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)

\(=\dfrac{2016}{2017}\)

\(A=\dfrac{1}{2}\left(\dfrac{2.2}{1.3}\right).\left(\dfrac{3.3}{2.4}\right)...\left(\dfrac{2020.2020}{2019.2021}\right)\)

\(=\dfrac{1.2.2.3.3...2020.2020}{1.2.2.3.3.4.4...2019.2021}\)

\(=\dfrac{1}{2021}\)

\(A=\dfrac{1}{2}\cdot\left(1+\dfrac{1}{1\cdot3}\right)\left(1+\dfrac{1}{2\cdot4}\right)\left(1+\dfrac{1}{3\cdot5}\right)...\left(1+\dfrac{1}{2019\cdot2021}\right)\)

\(A=\dfrac{1}{2}\left(1+\dfrac{1}{2^2-1}\right)\left(1+\dfrac{1}{3^2-1}\right)\left(1+\dfrac{1}{4^2-1}\right)...\left(1+\dfrac{1}{2020^2-1}\right)\)

\(A=\dfrac{1}{2}\cdot\dfrac{2^2}{\left(2-1\right)\left(2+1\right)}\cdot\dfrac{3^2}{\left(3-1\right)\cdot\left(3+1\right)}...\left(\dfrac{2020^2}{\left(2020-1\right)\cdot\left(2020+1\right)}\right)\)

\(A=\dfrac{1}{2}\cdot\dfrac{2}{1}\cdot\dfrac{2}{3}\cdot\dfrac{3}{2}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2020}{2019}\cdot\dfrac{2020}{2021}\)

\(A=\dfrac{1}{2}\cdot\dfrac{2}{1}\cdot\dfrac{3}{2}\cdot...\cdot\dfrac{2020}{2019}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2020}{2021}\)

\(A=\dfrac{1}{2}\cdot2020\cdot\dfrac{2}{2021}=\dfrac{2020}{2021}\)