# A curious summation question from pre-RMO 2016

This problem came in the Pre-RMO (Regional Mathematics Olympiad) - Delhi Region (INDIA). I've been solving these questions, and this particular summation question is proving to be quite difficult.

The question:

Consider the 50 term sums:

$$S = \frac{1}{1 \times 2} +\frac{1}{3\times 4} + \dots + \frac{1}{99 \times 100}$$ $$T = \frac{1}{51 \times 100} + \frac{1}{52 \times 99} + \dots + \frac{1}{100 \times 51}$$ The ratio $\frac{S}{T}$ is written in the lowest form $\frac{m}{n}$ where $m, n$ are relatively prime natural numbers. Find the value of $m + n$.

This is what I've tried so far:

We can re-write $S$ and $T$ as: $$S = \sum_{i=1}^{50}\frac{1}{(2i-1)\times(2i)}$$ $$= \sum_{i=1}^{50}\frac{1}{2i-1} - \frac{1}{2i}$$ $$T = \sum_{i=1}^{50}\frac{1}{(50+i)\times(101-i)}$$ $$= \frac{1}{151}\sum_{i=1}^{50}\frac{1}{50+i} + \frac{1}{101-i}$$

I expected these to be telescopic sums, and thus the terms could be cancelled out. But apparently not.

I've only reached this far. What are the next steps, or is there an easier way to find the ratio?

Thanks!

The answer is 153. Write $A=\sum_{i=1}^{50} \dfrac{1}{i}$ and $B=\sum_{i=1}^{50} \dfrac{1}{2i-1}$. Then $S=B-A/2$ and you can simplify $T$ as $T=\dfrac{2}{151}(B-A/2)$. Thus $S/T=151/2$.