Biến đổi thừa số tổng quát: \(1+\dfrac{1}{\left(k-1\right)\left(k+1\right)}\) \(=\dfrac{\left(k-1\right)\left(k+1\right)+1}{\left(k-1\right)\left(k+1\right)}\) \(=\dfrac{k^2}{\left(k-1\right)\left(k+1\right)}\).

Do đó \(1+\dfrac{1}{1.3}=\dfrac{2^2}{1.3}\), \(1+\dfrac{1}{2.4}=\dfrac{3^2}{2.4}\), \(1+\dfrac{1}{3.5}=\dfrac{4^2}{3.5}\),..., \(1+\dfrac{1}{2018.2020}=\dfrac{2019^2}{2018.2020}\), \(1+\dfrac{1}{2019.2021}=\dfrac{2020^2}{2019.2021}\). Từ đó suy ra \(\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)...\left(1+\dfrac{1}{2019.2021}\right)\) 

\(=\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.\dfrac{5^2}{4.6}.\dfrac{6^2}{5.7}...\dfrac{2019^2}{2018.2020}.\dfrac{2020^2}{2019.2021}\)

\(=\dfrac{2.2020}{2021}=\dfrac{4040}{2021}\)

Đặt A=\(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)+...+\(\dfrac{1}{n\left(n+1\right)\left(n+2\right)}\)

=>2A=\(\dfrac{2}{1.2.3}\)+\(\dfrac{2}{2.3.4}\)+...+\(\dfrac{2}{n\left(n+1\right)\left(n+2\right)}\)

=\(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\)\(\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\)

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

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

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

=>A=\(\dfrac{n^2+3n}{4n^2+12n+8}\)

program bai1;

var n,i:longint;s:real;

begin

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

s:=0;

for i:=1 to n do

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

writeln('Tong la ',s);

readln

end.

của Trương Quang Dũng thiếu uses crt;

ko khai báo s.

có 2 dấu ngoặc liên tiếp nên chương trình sẽ bị lỗi

1. \(A=\dfrac{2\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}{4\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}=\dfrac{2}{4}=\dfrac{1}{2}\)

2. \(B=\dfrac{1^2.2^2.3^2.4^2}{1.2^2.3^2.4^2.5}=\dfrac{1}{5}\)

3.\(C=\dfrac{2^2.3^2.\text{4^2.5^2}.5^2}{1.2^2.3^2.4^2.5.6^2}=\dfrac{125}{36}\)

4.D=\(D=\left(\dfrac{4}{5}-\dfrac{1}{6}\right).\dfrac{4}{9}.\dfrac{1}{16}=\dfrac{19}{30}.\dfrac{1}{36}=\dfrac{19}{1080}\)

hôi lì sít

bằng 0 nha bạn

tick cho mình

\(D=\left(1+\dfrac{1}{1.3}\right).\left(1+\dfrac{1}{2.4}\right)...\left(1+\dfrac{1}{2019.2021}\right)=\dfrac{4}{1.3}.\dfrac{9}{2.4}...\dfrac{2019.2021+1}{2019.2021}=\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}...\dfrac{2020.2020}{2019.2021}=\left(\dfrac{2}{1}.\dfrac{3}{2}...\dfrac{2020}{2019}\right).\left(\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2020}{2021}\right)=2020.\dfrac{2}{2021}=\dfrac{4040}{2021}\)

Lời giải:
Ta có: \(\frac{1}{k(k+1)(k+2)}=\frac{1}{2}.\frac{2}{k(k+1)(k+2)}=\frac{1}{2}.\frac{(k+2)-k}{k(k+1)(k+2)}\)

\(=\frac{1}{2}\left(\frac{k+2}{k(k+1)(k+2)}-\frac{k}{k(k+1)(k+2)}\right)=\frac{1}{2}\left(\frac{1}{k(k+1)}-\frac{1}{(k+1)(k+2)}\right)\)

Áp dụng vào bài toán:

\(\frac{1}{1.2.3}=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}\right)\)

\(\frac{1}{2.3.4}=\frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}\right)\)

\(\frac{1}{3.4.5}=\frac{1}{2}\left(\frac{1}{3.4}-\frac{1}{4.5}\right)\)

.......

\(\frac{1}{n(n+1)(n+2)}=\frac{1}{2}\left(\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\right)\)

\(\Rightarrow B=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{(n+1)(n+2)}\right)=\frac{1}{4}-\frac{1}{2(n+1)(n+2)}\)

\(B=\dfrac{1}{2}\left(\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+...+\dfrac{2}{n\left(n+1\right)\left(n+2\right)}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{n\cdot\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)\)

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

\(=\dfrac{1}{2}\cdot\dfrac{n^2+3n+2-2}{2\left(n+1\right)\left(n+2\right)}=\dfrac{n\left(n+3\right)}{4\left(n+1\right)\left(n+2\right)}\)

Kiểm tra lại đề xem thừa số cuối có đúng quy luật của dãy không.

1: \(S=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{101}{100}=\dfrac{101}{2}\)

2: \(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2006}{2007}=\dfrac{1}{2007}\)