Tính tổng S : (1+\(\dfrac{1}{1.3}\) ).(1+\(\dfrac{1}{2.4}\)).(1+\(\dfrac{1}{3.5}\))...(1+\(\dfrac{1}{2009.2011}\))
Tính tổng
\(S=\dfrac{1}{1.3}-\dfrac{1}{2.4}+\dfrac{1}{3.5}-\dfrac{1}{4.6}+\dfrac{1}{5.7}-\dfrac{1}{6.8}+\dfrac{1}{7.9}-\dfrac{1}{8.10}\)
\(S=\dfrac{1}{1.3}-\dfrac{1}{2.4}+\dfrac{1}{3.5}-\dfrac{1}{4.6}+\dfrac{1}{5.7}-\dfrac{1}{6.8}+\dfrac{1}{7.9}-\dfrac{1}{8.10}\)
\(S=\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}\right)-\left(\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+\dfrac{1}{8.10}\right)\)
\(S=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{7}-\dfrac{1}{9}\right)-\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{8}-\dfrac{1}{10}\right)\)
\(S=\dfrac{1}{2}-\dfrac{1}{18}-\dfrac{1}{4}+\dfrac{1}{20}\)
\(S=.C.A.S.I.O.\)
Viết thuật toán và chương trình để tính tổng \(A=\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+...+\dfrac{1}{n\left(n+2\right)}\)
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.
Tính :
(1 + \(\dfrac{1}{1.3}\) ) . ( 1+\(\dfrac{1}{2.4}\) ) . (1+\(\dfrac{1}{3.5}\)) . ... . ( 1+\(\dfrac{1}{2019.2021}\))
Lời giải:
Gọi tích trên là $A$
Xét thừa số tổng quát: $1+\frac{1}{n(n+2)}=\frac{n(n+2)+1}{n(n+2)}=\frac{(n+1)^2}{n(n+2)}$
Thay $n=1,2,3....,2019$ ta có:
$A=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}....\frac{2020^2}{2019.2021}$
$=\frac{2^2.3^2...2020^2}{(1.3)(2.4)(3.5)...(2019.2021)}$
$=\frac{(2.3....2020)(2.3...2020)}{(1.2.3...2019)(3.4...2021)}$
$=2020.\frac{2}{2021}=\frac{4040}{2021}$
Tính tổng A sau đây (n được nhập vào từ bàn phím):
A = \(\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+...+\dfrac{1}{1\left(n+2\right)}\)
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.
tính tổng
S=\(\dfrac{1}{1.3}-\dfrac{1}{2.4}+\dfrac{1}{3.5}-\dfrac{1}{4.6}+\dfrac{1}{5.7}-\dfrac{1}{6.8}+\dfrac{1}{7.9}-\dfrac{1}{8.10}\)
giúp nha
\(S=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+\dfrac{1}{7\cdot9}-\left(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+\dfrac{1}{8\cdot10}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}\right)-\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+\dfrac{2}{8\cdot10}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{9}\right)-\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{10}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{8}{9}-\dfrac{1}{2}\cdot\dfrac{2}{5}\)
\(=\dfrac{4}{9}-\dfrac{1}{5}\)
\(=\dfrac{11}{45}\)
Viết chương trình tính tổng B = \(\dfrac{1}{1.3}\) + \(\dfrac{1}{2.4}\) +\(\dfrac{1}{3.5}\) + ... + \(\dfrac{1}{n\left(n+2\right)}\)
________________________________________
Mọi người giúp mình với ạ, mình sắp thi rồi :((
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.
câu 1 tính
\(A=\dfrac{1}{2}\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)....\left(1+\dfrac{1}{2015.2017}\right)\)
\(A=\dfrac{1}{2}\left(2.\dfrac{2}{3}\right)\left(\dfrac{3}{2}.\dfrac{3}{4}\right)\left(\dfrac{4}{3}.\dfrac{4}{5}\right)....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)
\(=\dfrac{2016}{2017}\)
Thực hiện phép tính:
\(\left(1+\dfrac{1}{1.3}\right)+\left(1+\dfrac{1}{2.4}\right)+\left(1+\dfrac{1}{3.5}\right)+...+\left(1+\dfrac{1}{2019.2021}\right)\)
Sửa đề: A=(1+1/1*3)(1+1/2*4)*...*(1+1/2019*2021)
\(=\dfrac{2^2}{\left(2-1\right)\left(2+1\right)}\cdot\dfrac{3^2}{\left(3-1\right)\left(3+1\right)}\cdot...\cdot\dfrac{2020^2}{\left(2020-1\right)\left(2020+1\right)}\)
\(=\dfrac{2}{1}\cdot\dfrac{3}{2}\cdot...\cdot\dfrac{2020}{2019}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2020}{2021}=2020\cdot\dfrac{2}{2021}=\dfrac{4040}{2021}\)
Thực hiện phép tính:
\(\left(1+\dfrac{1}{1.3}\right)+\left(1+\dfrac{1}{2.4}\right)+\left(1+\dfrac{1}{3.5}\right)+...+\left(1+\dfrac{1}{2019.2021}\right)\)