the solution of the equation1!+2!+3!+.....x!=k^2 where k belongs to I are

(A)     1,3                                      (B)     2,3

(C)     4,5                                      (D)     -1,-3

Answer is 1 and 3

1!+2!+3!=9 which is a perfect square of 3 hence we get k = 3

1!=1 hence we get k = 1

so solution is k = 1,3

Further Discussion: Do any solutions exist other than these?

Sir there cannot be further any solution of this type of function because

1! + 2! + 3! +4! = 1 + 2 + 6 + 24 = 33  gives us end digit as 3 and further factorials have end digit as 0

5! = 120 , 6! = 720 ....

so if we keep on adding further factorials we will surely have digit at units place as 3

now there is no number whose square has end digit as 3 so we cannot have any more numbers other than 1 and 3

Tell me if i m wrong.

yes, that's correct