The sum(sequence) ends when the denominator is root 99 + root 100. I thought that if it continued on, it would be better to express the sum in an algebraic expression. The best alternative was to first think of sum of a sequence where nth term is 1/root (n) + root (n+1). To make it easier to calculate, I rationalized the fraction by multiplying root (n+1)-root(n), since (a+b)(a-b)=a^2-b^2. the denominator became {root (n+1)}^2-{root (n)}^2, which was equal to the equation n+1-n, which was 1. The numerator was 1 X {root (n+1)- root (n)}, and the whole fraction was rationalized to root(n+1)-root(n). We could now say that the sum was equal to the sum of the sequence root (n+1)-root (n) when n=1, n=2, ... n=n.
the sum is the same as -root(n)+ root (n+1) added up when n=1, n=2, ..n=n, which is
(-root 1+root 2)+(-root 2+root 3)+(-root 3+root 4)+...+{-root(n-1)+{root(n)}+ {-root(n)+root(n+1)}
From root 2 to -root n, everything gets cancelled out, leaving the -root 1 and + root(n+1) behind.
This shows that no matter how long the sequence (sum) is, we can simply substitute n into the expression. In this case, n=99, so the answer is
-root 1 + root 100
= -1+10
=9
I couldn't get the mathematical symbol for root, so please understand even if it is a little difficult to read. This may not be the best solution, but I hope it could be one of the methods for solving this question.