i)\(\dfrac{9}{10}\).\(\dfrac{23}{11}\) -\(\dfrac{1}{11}\) .\(\dfrac{9}{10}\) +\(\dfrac{9}{10}\)
=\(\dfrac{9}{10}\) \(\left(\dfrac{23}{11}-\dfrac{1}{11}\right)\) +\(\dfrac{9}{10}\)
=\(\dfrac{9}{10}\) .2+\(\dfrac{9}{10}\) =\(\dfrac{9}{10}\) (2+1)=\(\dfrac{9}{10}\) .3=\(\dfrac{27}{10}\)
j)A=\(\dfrac{2}{1.3}\) +\(\dfrac{2}{3.5}\) +...+\(\dfrac{2}{99.101}\)
A=1-\(\dfrac{1}{3}\) +\(\dfrac{1}{3}\) -\(\dfrac{1}{5}\) +...+\(\dfrac{1}{99}\) -\(\dfrac{1}{101}\)
A=1-\(\dfrac{1}{101}\) =\(\dfrac{100}{101}\)
k)A=\(\dfrac{4}{1.3}\) +\(\dfrac{4}{3.5}\) +...+\(\dfrac{4}{99.101}\)
A.\(\dfrac{1}{2}\)=\(\dfrac{2}{1.3}\)+\(\dfrac{2}{3.5}\)+...+\(\dfrac{2}{99.101}\)
A.\(\dfrac{1}{2}\)=1-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{5}\) +...+\(\dfrac{1}{99}\) -\(\dfrac{1}{101}\)
A.\(\dfrac{1}{2}\)=1-\(\dfrac{1}{101}\)=\(\dfrac{100}{101}\)
A=\(\dfrac{100}{101}\) :2=\(\dfrac{50}{101}\)
l)A=\(\dfrac{5}{3.7}\) +\(\dfrac{5}{7.11}\) +...+\(\dfrac{5}{81.85}\)
\(\dfrac{4}{5}\).A=\(\dfrac{4}{3.7}\) +\(\dfrac{4}{7.11}\) +...+\(\dfrac{4}{81.85}\)
\(\dfrac{4}{5}\).A=\(\dfrac{1}{3}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{11}\)+...+\(\dfrac{1}{81}\)-\(\dfrac{1}{85}\)
\(\dfrac{4}{5}\).A=\(\dfrac{1}{3}\)-\(\dfrac{1}{85}\)=\(\dfrac{88}{225}\)
A=\(\dfrac{88}{225}\):\(\dfrac{4}{5}\)=\(\dfrac{22}{45}\)