Sumn of nseries `(1+2)+(3+5)+(6+7)+(9+10)+............+(93+94)+(95=97)+(98+99)` will be :

The unit digit of the following expression (1 !)^(99) + (2!)^(98) + (3!)^(97) + (4!)^(96) + …..(99 !)^1 is :

The points (3,2,-4 ),( 5,4,-6 ) , (9,8,-10 )are

Let ((n),(k)) represents the combination of 'n' things taken 'k' at a time, then the value of the sum ((99),(97))+((98),(96))+((97),(95))+.........+((3),(1))+((2),(0)) equals-

1-2+3-4+........97-98 =?

[(1)/(1!99!)+(1)/(3!97!)+(1)/(5!95!)+......+(1)/(99!1!)]

The value of 99^(50) - 99.98^(50) + (99*98)/(1*2) (97)^(50) -…+ 99 is

The value of 99^(50) - 99.98^(50) + (99*98)/(1*2) (97)^(50) -…+ 99 is

The value of 99^(50) - 99.98^(50) + (99*98)/(1*2) (97)^(50) -…+ 99 is

The value of 99^(50) - 99.98^(50) + (99*98)/(1*2) (97)^(50) -…+ 99 is