\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)

\(=\frac{1}{3}+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+...+\left(\frac{1}{97}-\frac{1}{97}\right)-\frac{1}{99}\)

\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)

~ Hok tốt ~

\(\)

Viết thành 2 . (1/3.5 + 1/5.7 + 1/7.9 + ...+ 1/97.99

Tui hk bít nữa

a.  

\(M=1.\left[\frac{1}{3}-\frac{1}{5}+.....\frac{1}{97}-\frac{1}{99}\right]\)

\(M=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)

b.

\(N=\frac{3}{2}.\left[\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{197}-\frac{1}{199}\right]\)

\(N=\frac{3}{2}.\left[\frac{1}{5}-\frac{1}{199}\right]=\frac{291}{995}\)

mk đầu tiên nha bạn

\(M=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{97.99}\)

\(\Rightarrow M=\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\right)\)

\(\Rightarrow M=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(\Rightarrow M=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{99}\right)\)

\(\Rightarrow M=\frac{1}{2}.\frac{32}{99}\)

\(\Rightarrow M=\frac{16}{99}\)

\(M=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)

\(M=\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)

\(M=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5.}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(M=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{99}\right)=\frac{16}{99}\)

nhân M với 2  

c/m theo công thức ta được

10M=1/3-1/99

10M=32/99

M=16/495

dung thi tk nhâ

Ta có S=2/3+2/3.5+2/5.7+2/7.9+...+2/97.99 
           =2/3+1/3-1/5+1/5-1/7+1/7-1/9+...+1/97-1/99
           =2/3+1/3+(1/5-1/5)+(1/7-1/7)+...+(1/97-1/97)+1/99
           =1+0+0+0+...+0+1/99
           =1+1/99
           =100/99
Mà 100/99>1.Suy ra S>1
   Vậy S>1
            

S=1-1/3 + 1/3 - 1/5 + ... + 1/97 - 1/99

=1 - 1/99 => S<1

M=(1/3-1/5)+(1/5+1/7)+...+(1/97+1/99)

M=1/3+(1/5-1/5)+...+(1/97-1/97)-1/99

M=1/3-1/99

M=32/99

\(\frac{32}{99}\)

Đây là bài toán tìm tổng dãy số có quy luật.

Để ý thấy rằng \(\frac{1}{n\left(n+2\right)}=\frac{1}{2}.\frac{2}{n\left(n+2\right)}=\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right)\)

Vậy thì \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{n\left(n+2\right)}=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n+2}\right)\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{n+2}\right)=\frac{5}{36}\Rightarrow\frac{1}{3}-\frac{1}{n+2}=\frac{5}{18}\)

\(\Rightarrow\frac{1}{n+2}=\frac{1}{18}\Rightarrow n=16.\)

\(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{n\left(n+2\right)}=\frac{5}{36}\)

\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{n}-\frac{1}{n+2}=\frac{5}{36}\)

\(\frac{1}{3}-\frac{1}{n+2}=\frac{5}{36}\)

\(\frac{12}{36}-\frac{1}{n+2}=\frac{5}{36}\)

\(\frac{1}{n+2}=\frac{7}{36}\)

\(\Rightarrow\frac{7}{7\left(n+2\right)}=\frac{7}{36}\)

\(7\left(n+2\right)=36\)

n + 2 = 36/7

n = 36/7 - 2

( Tự tính KQ nha )

1/3.5+1/5.7+1/7.9+....+1/n(n+2)=5/36

=1(1/3.5+1/5.7+1/7.9+....+1/n(n+2)

=1/2(2/3.5+2/5.7+2/7.9+....+2/n(n+2)

=1/2(1/3-1/5+1/5-1/7+1/7-1/9+...+1/n-1/n+2

=1/2(1/3+-1/5+1/5+-1/7+1/7+-1/9+...+1/n+-1/n+2

=1/2(1/3+0+0......+-1/n+2)=5/36

=1/2(1/3+1/n+2)=5/36

=(1/3+1/n+2=5/36:1/2

=1/3+1/n+2=5/18

=1/n+2=5/18-1/3

1/n+2=-1/18

=1/n+2=1/-18

=1/n=1/-18-2

=1/n=1/-20

n=-20

=

= \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\)

\(=\dfrac{1}{3}-\dfrac{1}{99}=\dfrac{32}{99}\)

= \(\dfrac{\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}}{2}\)

= \(\dfrac{\dfrac{1}{3}-\dfrac{1}{99}}{2}=\dfrac{16}{99}\)

Bài kia mk nhầm sorry