\(I=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2009.2010}\)

\(I=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2009}-\frac{1}{2010}\)

\(I=1-\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+.....+\left(\frac{1}{2009}-\frac{1}{2009}\right)-\frac{1}{2010}\)

\(I=1-0-0-...-0-\frac{1}{2010}\)

\(I=1-\frac{1}{2010}=\frac{2009}{2010}\)

I = 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/2009.2010

I = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/2009 - 1/2010

I = 1 - 1/2010

I = 2009/2010 

Vậy I = 2009/2010

= 1 - 1/2 . 1/2 -1/3 . 1/3 - 1/4 ... 1/2009 - 1/2010

= 1 - 1/ 2010

=1/2010

1/1.2+1/2.3+1/3.4+...+1/2009.2010

=1-1/2+1/2-1/3+...+1/2009-1/2010

=1-1/2010

=2009/2010

=(1-1/2)+(1/2-1/3)+...+(1/9-1/10) 
=1-1/10 
=9/10

I=1-1/2+1/2-1/3+1/3-1/4+...+1/2009-1/2010

I=1-1/2010

I=2009/2010

Vậy I=2009/2010

I = 1/1-1/2+1/2-1/3+1/3-1/4+...+1/2009-1/2010

I = 1-1/2010

I = 2009/2010

Chúc bạn học tốt nha

ta thấy: 1/1 - 1/2 = 1/2 = 1/1.2

            1/2 - 1/3 = 1/6 = 1/2.3

            1/3 - 1/4 = 1/12 = 1/3.4

tớ nêu cách làm rùi đó

\(I=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2009}-\frac{1}{2010}\)

\(I=1-\frac{1}{2010}\)

\(I=\frac{2009}{2010}\)

Tại sao ra kết quả như vậy hả bạn ???

=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6

=1/1+-1/2+1/2+-1/3+1/3+-1/4+1/4+-1/5+1/5+-1/6

=1/1+-1/6=5/6

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)

\(=1-\frac{1}{6}=\frac{5}{6}\)

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}\)

=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}\)

=\(1-\dfrac{1}{5}\)

=\(\dfrac{4}{5}\)

1/1 - 1/2+ 1/2 - 1/3 + 1/3 - 1/4+ 1/4 - 1/5

 = 1/1 - 1/5

= 4/5

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}=1-\dfrac{1}{5}=\dfrac{4}{5}\)

Khônh hiểu

437/60

Ta có : \(\frac{1}{1.2}\)+  \(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+\(\frac{1}{4.5}\)+\(\frac{1}{5.6}\)+\(\frac{1}{6.7}\)

= 1  - \(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+\(\frac{1}{4}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{6}\)+\(\frac{1}{6}\)-\(\frac{1}{7}\)

= 1 - \(\frac{1}{7}\)=  \(\frac{6}{7}\)

=1-1/2+1/2-1/3+1/3-1/4+...+1/6-1/7=1-1/7=6/7

1/1 -1/2 +1/2 -1/3 +1/3 -1/4 +1/4 -1/5 +1/5 -1/6 +1/6 -1/7

1/1- 1/7 =6/7

Ta có : 3A = 1.2.3 + 2.3.3 + 3.4.3 + .... + n.( n + 1 ).3

=> 3A = 1.2.( 3 - 0 ) + 2.3.( 4 - 1 ) + 3.4.( 5 - 2 ) + ..... + n.( n + 1 ).[ ( n + 2 ) - ( n - 1 ) ]

=> 3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ..... + n.( n + 1 ).( n + 2 ) - ( n - 1 ).n.( n + 1 )

=> 3A = ( 1.2.3 - 1.2.3 ) + ( 2.3.4 - 2.3.4 ) + .... + [ ( n - 1 ).n.( n + 1 ) - ( n - 1 ).n.( n + 1 ) ] + n.( n + 1 ).( n + 2 )

=> 3A = n.( n + 1 ).( n + 2 )

=> A = \(\frac{n.\left(n+1\right).\left(n+2\right)}{3}\)