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 

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

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 !

\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)

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

\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{101}\right)=\frac{1}{2}.\frac{98}{303}=\frac{49}{303}\)

49/303

tich bù 4 điểm nữa là tớ tròn 60 rồi 

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

\(=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

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

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