Question

a=1/1.2+1/2.3+1/3.4+.......1/99.100

 

Answer 1

$\frac{2a}{b.(b+a).(b+2a)}=\frac{1}{b.(b+a)}-\frac{1}{(b+a).(b+2a)}$
$\frac{3a}{b.(b+a).(b+2a).(b+3a)}=\frac{1}{b.(b+a).(b+2a)}-\frac{1}{(b+a)(b+2a)(b+3a)}$  

Answer 2

xin lỗi nhiều nhé chờ chút mik làm lại

\(\frac{2a}{b.\left(b+a\right).\left(b+2a\right)}=\frac{1}{b\left(b+a\right)}=\frac{1}{\left(b+a\right)\left(b+2a\right)}.\)

\(\frac{3a}{b\left(b+a\right)\left(b+2a\right)\left(b+3a\right)}=\frac{1}{b\left(b+a\right)\left(b+2a\right)}-\frac{1}{\left(b+a\right)\left(b+2a\right)\left(b+3a\right)}.\)

tích nha

Answer 3

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{99.100}\)

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

\(=1-\frac{1}{100}=\frac{99}{100}\)