\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{38.40}\)

=\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{38}-\frac{1}{40}\)

=\(\frac{1}{2}-\frac{1}{40}\)

=\(\frac{19}{40}\)

= 2 *[1/2 * 1/4 +1/4 *1/6 +1/6*1/8+...+1/38*1/40

=2*[1/2 - 1/40]

=2 * (-19/40)

= -380

=1/2 - 1/4 + 1/4 - 1/6+1/6-1/8+1/8-...-1/38+1/38-40

=1/2+0+0+0+...+0-1/40

=1/2-1/40

=20/40-1/40

=19/40

A = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +...+ 1/2011 - 1/2012

A = 1 - 1/2012

A = 2011/2012

B = 1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 +...+ 1/2010 - 1/2012

B = 1/2 - 1/2012

B = 1005/2012

a) \(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2011\cdot2012}\)

\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2011}-\dfrac{1}{2012}\)

\(A=1-\dfrac{1}{2012}\)

\(A=\dfrac{2011}{2012}\)

b) \(B=\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+...+\dfrac{1}{2010\cdot2012}\)

\(B=\dfrac{1}{2}\cdot\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{2010\cdot2012}\right)\)

\(B=\dfrac{1}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2010}-\dfrac{1}{2012}\right)\)

\(B=\dfrac{1}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{2012}\right)\)

\(B=\dfrac{1}{2}\cdot\dfrac{1005}{2012}\)

\(B=\dfrac{1005}{4024}\)

Số hạng thứ 50 của dãy là: \(\frac{1}{100.102}\)

Tổng 50 số hạng đầu của dãy là:\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+.....+\frac{1}{100.102}=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{100}-\frac{1}{102}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{102}\right)=\frac{1}{2}.\frac{25}{51}=\frac{25}{102}\)

phân số thứ 50 là 1/98.100

1/2.4+1/4.6+1/6.8+.......+1/98.100

=2.(1/2-1/4+1/4-1/6+1/6-1/8+.........+1/98-1/100).1/2

=(1-1/2+1/2-1/3+1/3-1/4+...........+1/49-1/50).1/2

=(1-1/50).1/2

=49/50.1/2

=49/100

Số hạng thứ 50 theo quy luật là: \(\frac{1}{100.102}\)

Gọi tổng 50 số hạng đầu là S

Ta có: \(S=\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{100.102}\)

\(\Leftrightarrow2S=\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{100.102}\)

\(\Leftrightarrow2S=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{100}-\frac{1}{102}=\frac{1}{2}-\frac{1}{102}=\frac{25}{51}\)

\(\Rightarrow S=\frac{25}{51}:2=\frac{25}{102}.\)

Bạn Don''t look at me làm đúng rồi ấy

A=1/2.4+1/4.6+........+1/100.102

A=1/2-1/4+1/4-1/6+.......+1/100-1/102

A=1/2-1/102

A=51/102-1/102

A=50/102

A=25/51

\(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{18}-\frac{1}{20}\)

\(A=\frac{1}{2}-\frac{1}{20}\)

\(A=\frac{10}{20}-\frac{1}{20}\)

\(A=\frac{9}{20}\)

\(S=\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{2018.2020}\)

\(S=\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2018.2020}\right)\)

\(S=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2018}-\frac{1}{2020}\right)\)

\(S=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2020}\right)\)

Tự tính

S=1/2.4+1/4.6+1/6.8+...+1/2018.2020

S=1/2.(2/2.4+2/4.6+2/6.8+...+2/2018.2020)

S=1/2.(1-1/4+1/4-1/6+1/6-1/8+...+1/2018-1/2020)
S=1/2.(1-1/2020)

S=1/2.(2020/2020-1/2020)

S=1/2.2019/2020

S=2019/4040

\(S=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2018}-\frac{1}{2020}\right)\)

\(S=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2020}\right)\)

\(S=\frac{1}{4}-\frac{1}{4040}=\frac{1010}{4040}-\frac{1}{4040}=\frac{1009}{4040}\)