Ta có \(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+...+\dfrac{1}{2n\left(2n+2\right)}=\dfrac{1009}{4038}\)

\(\Leftrightarrow\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{2n\left(2n+2\right)}=\dfrac{1009}{2019}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2n}-\dfrac{1}{2n+2}=\dfrac{1009}{2019}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2n+2}=\dfrac{1009}{2019}\)

\(\Leftrightarrow\dfrac{n}{2n+2}=\dfrac{1009}{2019}\)

\(\Leftrightarrow2019n=1009\left(2n+2\right)\)

\(\Leftrightarrow2019n=2018n+2018\)

\(\Leftrightarrow n=2018\)

\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{2n.\left(2n+2\right)}\))

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

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

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

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

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

bạn ơi mình ko hiểu chỗ \(\frac{1}{4}-\frac{1}{2.\left(2n+2\right)}\)

tự làm là hạnh phúc của mỗi công dân.

xét          \(VT=\frac{2}{2}\left(\frac{1}{2.4}+\frac{1}{4.6}+......+\frac{1}{2n.\left(2n+2\right)}\right)\)     (1)

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

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

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

\(=\frac{1}{4}-\frac{1}{4n+4}\)

mà theo bài ra   (1) = \(\frac{502}{2009}\)

<=>\(\frac{1}{4}-\frac{1}{4n+4}=\frac{502}{2009}\)

<=>\(\frac{1}{4n+4}=\frac{1}{4}-\frac{502}{2009}\)

<=>\(\frac{1}{4n+4}=\frac{1}{8036}\)

<=> 4n+4=8036

<=> 4n=8032

<=> n=2008

=) \(\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2n\left(2n+2\right)}\right)=\frac{502}{2009}\)
=) \(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2n}-\frac{1}{2n+2}\right)=\frac{502}{2009}\)
=) \(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2n+2}\right)=\frac{502}{2009}\)
=) \(\frac{1}{2}-\frac{1}{2n+2}=\frac{502}{2009}:\frac{1}{2}=\frac{1018}{2009}\)
=) \(\frac{1}{2n+2}=\frac{1}{2}-\frac{1018}{2009}=\frac{-27}{4018}\)
=) \(\frac{-1}{-\left(2n+2\right)}=\frac{-27}{4018}\)
=) \(\frac{-27}{27.-\left(2n+2\right)}=\frac{-27}{4018}\)
=) \(27.-\left(2n+2\right)=4018\)
=) \(-\left(2n+2\right)=4018:27=\frac{4018}{27}\)
=) \(2n+2=\frac{-4018}{27}\)
=) \(2n=\frac{-4018}{27}-2=\frac{-4072}{27}\)
=) \(n=\frac{-4072}{27}:2=\frac{-2036}{27}\)
\(\)
 

A = 1.100 + 2.99 + 3.98 + 98.3 + 99.2 + 100.1

1.100 = 1.100 = 1.100

2.99 = 2.(100 - 1) = 2.100 - 1.2

3.98 = 3.(100 - 2) = 3.100 - 2.3

4.97 = 4.(100 - 3) = 4.100 = 3.4

...............................................................

100.1 = 100.(100 - 99) = 100.100 - 99.100

Cộng vế với vế ta có:

A = 1.100+2.100+...+99.100+100.100 - (1.2 +2.3+ 3.4+...+99.100)

Đặt B = 1.100 + 2.100+...+99.100 + 100.100

C = 1.2 + 2.3 + 3.4 +...+ 99.100

A = B - C

B = 1.100 + 2.100 + ...+ 99.100 + 100.100

B = 100.(1+ 2+ ... + 99+ 100)

B = 100.(100 + 1) x 100 : 2

B = 505000

C = 1.2 + 2.3 + 3.4 +...+ 99.100

3C = 1.2.3 + 2.3.3 +..+99.100.3

1.2.3 = 1.2.3

2.3.3 = 2.3.(4 - 1) = 2.3.4 - 1.2.3

99.100.3 = 99.100.(101 - 98)=99.100.101-98.99.100

Cộng vế với vế ta có:

3C = 99.100.101

C = 99.100.101 : 3

C = 333300

A = B - C

A = 505000 - 333300

A = 171700

Câu b:

A = 9+99+ 999+...+9999...99(1000 chữ số 9)

9 = - 1 + 10

99 = - 1 + 100

999 = - 1 + 1000

...............................

999...999 = -1 + 1000...00(1000 chữ số 0)

Cộng vế với vế ta có:

B = - 1 x 1000 + 11111...10(1000 chữ số 1)

B = 111....110110 (999 chữ số 1)