Thai maths exam (PAT1)

Algebra Level 5

\[A=\frac{1}{1\cdot 2}+\frac{1}{3\cdot 4}+\frac{1}{5\cdot 6}+...+\frac{1}{2015\cdot 2016}\] \[B=\frac{1}{1009\cdot 2016}+\frac{1}{1010\cdot 2015}+\frac{1}{1011\cdot 2014}+...+\frac{1}{2016\cdot 1009}\] \[\frac{20A}{11B}= \ ?\]

×

Problem Loading...

Note Loading...

Set Loading...