Evaluating 1/(1+2) + 1/(1+2+3) + 1/(1+2+3+4)... + 1/(1+2+3+4...+100)

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Evaluate 1/(1+2) + 1/(1+2+3) + 1/(1+2+3+4)... + 1/(1+2+3+4...+100)

This is a math competition question that I have tried solving for a long time as I can't use a calculator. What is the simplest way of solving this question without using a calculator?

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Hint:

$$\dfrac1{\sum_{r=1}^nr}=\dfrac2{n(n+1)}=\dfrac{2(n+1-n)}{n(n+1)}=f(n)-f(n+1)$$ where $f(r)=\dfrac2r$

Use Telescoping series