Question

B=\(\dfrac{1}{6x10}\)+\(\dfrac{1}{10x14}\)+.....+\(\dfrac{1}{402x406}\)
 

Answer 1

\(B=\dfrac{1}{6\cdot10}+\dfrac{1}{10\cdot14}+...+\dfrac{1}{402\cdot406}\)

\(=\dfrac{1}{4}\left(\dfrac{1}{6}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{14}+...+\dfrac{1}{402}-\dfrac{1}{406}\right)\)

\(=\dfrac{1}{4}\cdot\dfrac{400}{2436}=\dfrac{100}{2436}=\dfrac{25}{609}\)