\hspace{-16}$If $\mathbf{[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+.......+[\sqrt{x^2-1}]=y}$\\\\ Then $\mathbf{(x,y)=}$
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1057in this question we have to observe that difference of consecutive perfect squares are in AP bcoz (x+1)2 -x2=2x+1
therefore in y 1 will occur thrice.....2 five times ,3 7 times and so on...and the last term will be added 1 less...
therefore y is the sum of the series whose nth term is (n-1)(2x-1)
therefore using summation formulas....we get...
y=x(x+1)(4x-7) 6+x
