1/3+1/15+1/35+1/63+1/99+……+1/9999
=1/(1×3)+1/(3×5)+1/(5×7)+1/(7×9)+1/(9×11)+……+1/(99×101)
=1/2(1-1/3)+1/2(1/3-1/5)+1/2(1/5-1/7)+1/2(1/7-1/9)+1/2(1/9-1/11)+……+1/2(1/99-1/101)
=1/2(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+……+1/99-1/101)
=1/2(1-1/101)
=1/2×(100/101)
=50/101

1/3+1/15+1/35+1/63+1/99+……+1/9999

=1/(1×3)+1/(3×5)+1/(5×7)+1/(7×9)+1/(9×11)+……+1/(99×101)

=1/2(1-1/3)+1/2(1/3-1/5)+1/2(1/5-1/7)+1/2(1/7-1/9)+1/2(1/9-1/11)+……+1/2(1/99-1/101)

=1/2(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+……+1/99-1/101)

=1/2(1-1/101)

=1/2×(100/101)

=50/101 

T nghĩ đề là phép + chứ nhỉ?! phép trừ thì s lm đc?!

\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+...+\dfrac{1}{9999}\)

\(=\dfrac{1}{2}+\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{99\cdot101}\)

\(=\dfrac{1}{2}+\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{99\cdot101}\right)\)

\(=\dfrac{1}{2}+\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=\dfrac{1}{2}+\dfrac{1}{2}\left(1-\dfrac{1}{101}\right)=\dfrac{1}{2}+\dfrac{1}{2}\cdot\dfrac{100}{101}=\dfrac{201}{202}\)

p/s: Nghĩ vậy còn đề là trừ thì ~~ Chịu ~~

\(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{15}-\dfrac{1}{35}-\dfrac{1}{63}-...-\dfrac{1}{9999}\)

\(=\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+...+\dfrac{1}{9999}\right)\)

\(=\dfrac{1}{2}-\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{99.101}\right)\)

\(=\dfrac{1}{2}-\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=\dfrac{1}{2}-\dfrac{1}{2}\left(1-\dfrac{1}{101}\right)\)

\(=\dfrac{1}{2}-\dfrac{1}{2}.\dfrac{100}{101}\)

\(=\dfrac{1}{2}-\dfrac{50}{101}\)

\(=\dfrac{1}{202}.\)

Tui nghĩ là 49/303 

49/303 chắc chắn lun mình giải rùi tick nha

cô hiệu phó trường mình chữa đó.

1/(3x5) + 1/(5x7) + 1/(7x9) + 1/(9x11)+...  + 1/(99x101)

(1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+...+1/99-1/101) : 2

(1/3 - 1/101) : 2 =  98/303  :  2

49/303

Bạn đưa về dãy tổng

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

Có thể tính nhanh vì đây là dãy đặc biệt 

\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{9999}\)

= \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{99.101}\)

= \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{99}-\frac{1}{101}\)

Sau khi lược bỏ các phân số ( phân số cộng với nhau bằng 0 coi như là không cộng)

Ta còn : \(\frac{1}{3}-\frac{1}{101}\)=\(\frac{98}{303}\)

Đáp số: \(\frac{98}{303}\)

\(\frac{98}{303}\)

A = 1/15 + 1/35 + 1/ 63 + 1/99 + ...+ 1/9999

A     = 1/(3x5) + 1/(5x7) + 1/(7x9) + 1/(9x11) + ... + 1/(99 x 101)

Ax2 = 2/(3x5) + 2/(5x7) + 2/(7x9) + 2/(9x11) + ... + 2/(99 x 101)

Ax2 = 1/3 – 1/5 + 1/5 – 1/7 + 1/7 – 1/9 + 1/9 – 1/11 + ...+ 1/99 – 1/101

Ax2 = 1/3 – 1/101 = 98/303

A    =  98/303 : 2

A    =  49/303

49/303

nếu muốn giải hẳn ra thì phải tick 2 lần đó nhe !