\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{101}\right)=\frac{1}{2}\times\frac{98}{303}=\frac{49}{303}\)
mình gợi ý là dạng chuỗi, tích mình nha rồi mình giải cho
đặt biểu thức là A nhé!
A=1/15+1/35+1/63+1/99+...+1/9999
=>2A=2 x (1/15+1/+35+1/63+1/99+...+1/9999)
2A=2/15+2/35+2/63+2/99+...+2/9999
2A=1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+....+1/99-1/101
2A=1/3-1/101
2A=98/303
=>A=98/303:2
A=49/303
\(\frac{1}{15}\)+ \(\frac{1}{35}\)+ \(\frac{1}{63}\)+ \(\frac{1}{99}\)+ .... + \(\frac{1}{9999}\)
= \(\frac{1}{3\text{x}5}\)+ \(\frac{1}{5\text{x}7}\)+ \(\frac{1}{7\text{x}9}\)+ \(\frac{1}{9\text{x}11}\)+ .... + \(\frac{1}{99\text{x}101}\)
= ( \(\frac{1}{3}\)- \(\frac{1}{5}\)+ \(\frac{1}{5}\)- \(\frac{1}{7}\)+ \(\frac{1}{7}\)- \(\frac{1}{9}\)+ \(\frac{1}{9}\)- \(\frac{1}{11}\)+ ... + \(\frac{1}{99}\)- \(\frac{1}{101}\)) : 2
= ( \(\frac{1}{3}\)- \(\frac{1}{101}\)) : 2
= ( \(\frac{101}{303}\)- \(\frac{3}{303}\)) : 2
= \(\frac{98}{303}\): 2
= \(\frac{98}{606}\)= \(\frac{49}{303}\)
= ( \(\frac{1}{3}\)-
