\(S=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\)

\(=1-\dfrac{1}{101}=\dfrac{100}{101}\)

NHANH GIÚP TỚ VỚI 

\(D=\frac{1}{3}\cdot\frac{3}{5}\cdot\frac{5}{7}\cdot\frac{7}{9}\cdot...\cdot\frac{99}{101}\)

\(D=\frac{1\cdot3\cdot5\cdot7\cdot...\cdot99}{3\cdot5\cdot7\cdot9\cdot...\cdot101}\)

\(D=\frac{1}{101}\)

`A=1/(1xx3)+1/(3xx5)+1/(5xx7)+...+1/(95xx97)+1/(97xx99)`

`=> 2A=2/(1xx3)+2/(3xx5)+...+2/(97xx99)`

`=> 2A=1-1/3+1/3-1/5+...+1/97-1/99`

`=> 2A=1-1/99`

`=> 2A=98/99`

`=> A=49/99`

Đặt A = 1 x 3 + 3 x 5 + 5 x 7 + ....+ 97 x 99

=> 6A = 1 x 3 x 6 + 3 x 5 x 6 + 5 x 7 x6 + ...+ 97 x 99 x 6

6A = 1 x 3 x ( 5+1) + 3 x 5 x (7-1) + 5 x 7 x ( 9-3) + ...+ 97 x 99 x (101- 95)

6A = 1 x 3 x 5 + 1 x 3 + 3 x 5 x 7 - 1x3x5 + 5 x 7 x 9 - 3x5x7 + ...+ 97x99x101 - 95x97x99

6A = ( 1 x 3 + 1 x 3 x 5 + 3 x 5 x 7 + 5x7x9+...+97x99x101) - (1x3x5+3x5x7+...+95x97x99)

6A = 1 x 3 + 97x99x101

\(A=\frac{1x3+97x99x101}{6}=161651\)

11 + báo cáo

=1/2 - 1/101

Sửa đề :

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

\(1-\frac{1}{101}=\frac{100}{101}\)

= 3/2x(1/3-1/5) + 3/2x(1/5 -1/7) +......+ 3/2x(1/97-1/99)

=3/2x(1/3-1/5+1/5-1/7+1/7-1/9+....+1/97-1/99) = 3/2x(1/3-1/99)

=3/2x32/99=(3x32)/(2/99) = 16/33

chúc bạn học tốt!

\(=\frac{3}{2}\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{99}-\frac{1}{100}\right).\)

\(=\frac{3}{2}.\left(\frac{1}{1}-\frac{1}{100}\right)=\frac{3}{2}.\frac{99}{100}=\frac{297}{200}\)

Cảm ơn bạn