at n=3 , Qwould be 9. Is there an analytical soution for this?

The answer is Option 2). For n=1,1!=1 is a perfect square.For n=3,Q=1!+2!+3!=1+2+6=9,a perfect square.Therefore,Q= perfect square for n=1,3.  

answer is Option 2.

The Condition is satisfied @ n = 1, 3...

@ VINAY, yes there is analytical Soln to this problem also...

Soln.  

for n =1  Q =1

for n = 2 Q = 1 +2 =3

for n = 3 Q = 1 +2 + 6 = 9

for n = 4 Q = 1 +2 + 6 + 24 = 33

for n = 5 Q = 1 +2 + 6 + 24 + 120 = 153

for all numbers 5 onwards... unit digit will not change...means it will remain 3... But there can not be a perfect square with unit digit = 3.

Hence, for all n>=5, Q can never be a perfect square.

So, we have only 2 values of n i.e 1 and 3

I hope this soln. is most optimal... If anyone have a better approach please update me..

Thanks

   That s a smart approach! Thank you.

This is one of the problem given in one of the Quant Test. Nice approach.

the answer is 2...........

m also use same approch

thanks for explanation!