The sequence $x_n$ is defined by $x_{n+1} = x_n^2 + x_n$. Suppose S = $$\sum_{i=1}^{99} {1\over x_i+1}$$ If $x_1 = {1\over 60}$ , find $[S]$ (where $[x]$ represents integral part of $x$)
MY WORK :- Well basically we see that $x_{n+1} = x_n (x_n+1)$ thus we get that ${x_{n+1}\over x_n} = x_n+1$. Thus ${1\over x_n+1} = {x_n\over x_{n+1}}$. So basically the series sums up to ${x_1\over x_2}+{x_2\over x_3}+{x_3\over x_4}+\cdots + {x_{n-1}\over x_n}+{x_n\over x_{n+1}}$. But this is where i am stuck; what to do now; how to simplify ??