Question

giup mk bai nai voi

Chứng tỏ rằng

\(\frac{k}{n.\left(n+k\right)}\)=\(\frac{1}{n}-\frac{1}{n+k}\)    

Aps dụng;Tính; S=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)

Answer 1

= 2 x [1 - 1/3 + 1/3 - 1/5 + 1/5 -1/7 +1/7 -1/9 + .., +1/99 - 1/101

= 2 x [ 1 - 1/101 ]

= 2 x 100/101

= 200/101

t cho mik nha

Answer 2

   \(\frac{2}{1.3}\)+\(\frac{2}{3.5}\)+\(\frac{2}{5.7}\)+.........+\(\frac{2}{99.101}\)

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

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

Answer 3

= 2 x [1 - 1/3 + 1/3 -1/5 + 1/5 -1/7 +1/7 - 1/9 +...+ 1/99 - 1/101]

= 2 x [ 1 - 1/101 ]

= 2 x 100/101

= 200/101