There are $$100$$ points $$P_1,P_2,\ldots,P_{100}$$ on a line such that the distance between $$P_i$$ and $$P_{i+1}$$ is $$\frac{1}{i}$$ for $$1\leq i\leq99$$. The sum of the distances between every pair of points is
Correct Answer: 4950
Solution
The gap $$\frac{1}{i}$$ occurs in the distance between $$P_r$$ and $$P_s$$ exactly when $$r\leq i<s$$. There are $$i(100-i)$$ such pairs, so this gap contributes $$\frac{1}{i}\mathbin{\times}i(100-i)=100-i$$. Summing over $$i=1$$ to $$99$$ gives $$99+98+\cdots+1=4950$$.
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