The sum `[1/(1!99!)+1/(3!97!)+1/(5!95!)+.......+1/(99!1!)]` is equal to

sum_(n=1)^(99) n! (n^2 + n+1) is equal to :

[[-(1)/(3)]+[-(1)/(3)-(1)/(100)]+[-(1)/(3)-(2)/(100)]+...+[-(1)/(3)-(99)/(100)] is equal to (where [.] denotes greatest integer function)

[-(1)/(3)]+[-(1)/(3)-(1)/(100)]+[-(1)/(3)-(2)/(100)]+….+[-(1)/(3)-(99)/(100)] is equal to (where [.] denotes greatest integer function)

The sum of 1/(sqrt(2)+1) + 1/(sqrt(3) + sqrt(2)) + 1/(sqrt(4) + sqrt(3)) +.....1/(sqrt(100) + sqrt(99)) is equal to:

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

The value of sqrt(97 xx 98 xx 99 xx100+1) is equal to

Conisder x = 4 tan^(-1)((1)/(5)),y = tan^(-1)((1)/(70))and z = tan^(-1)((1)/(99)) . What is x equal to ?