xk+1 = xk(xk+1)
1xk+1 = 1xk(xk+1) = 1xk - 1xk+1
or,1xk+1 = 1xk - 1xk+1
The required sum is =
1x1+1 + 1x2+1 + 1x3+1 + ..........+1x100+1
= 1x1-1x2+........+1x99-1x100
=1x1 - 1x100
x3 >1
If xk >1 then xk+1 = xk(xk+1)>1
Using induction we can say that x100 >1
Therefore required sum = 1x1 - 1x100 = 2 - 1x100.
Since x100 >1 , 1x100<1
Therefore the integer part of the sum is = 1.