If [x] is the greater integer less than or equal to ‘x’, then find the value of the following series $$[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+[\sqrt{4}]+....+[\sqrt{360}]$$

$$[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+[\sqrt{4}]+....+[\sqrt{360}]$$ = 1 + 1 + 1 + 2 + 2 + 2 + 2 + 2 + 3 + 3 + 3.. 7 times + 4 + 4 +......9 times +.........+ 18 + 18 + ......37 times

= 1*3 + 2*5 + 3*7 + 4*9+............+18*37 

= $$\sum_1^{18}$$ n(2n+1)

=  $$\sum_1^{18}$$ $$2n^2+n$$

= $$\frac{2n*(n+1)*(2n+1)}{6}$$ + $$\frac{n*(n+1)}{2}$$

= $$\frac{2*18*(19)*(37)}{6}$$ + $$\frac{18*19}{2}$$

= 4389

Therefore our answer is option 'A'