1

I am trying to compute the sum of these fractions:

$$\frac{3}{1}+\frac{3}{1+2}+\frac{3}{1+2+3} + \dots + \frac{3}{1+2+3+\dots+100}.$$

I believe the denominators are a triangular number sequence, therefore the expression should convert to

$$3\Bigl(\frac{2}{1×2}+\frac{2}{2×3}+\frac{2}{3×4} + \dots +\frac{2}{100×101}\Bigr)$$

then, this can then be converted to

$$6\biggl[\Bigl( \frac{1}{1}-\frac{1}{2}\Bigr) + \Bigl(\frac{1}{2}-\frac{1}{3}\Bigr) + \Bigl(\frac{1}{3}-\frac{1}{4}\Bigr) + \dots + \Bigl(\frac{1}{100}-\frac{1}{101}\Bigr)\biggr]$$

then, this simplifies to $6\Bigl(\frac{1}{1}$ - $\frac{1}{101}\Bigr)$

with the final answer being $\mathbf{6\frac{100}{101}}$

However, the answer key we obtained specifies the answer as $5\frac{95}{101}$

I would like to have some help on finding out if I made any errors with my solution hence our answer not being the same as the $5\frac{95}{101}$ given by the answer key? Thank you in advance.

Bernard
  • 175,478
RC Wong
  • 57

1 Answers1

1

Call the $n$th Triangular number $T_n$. Then, noting $T_{100}=5050$, the reciprocal sum of the first $100$ triangular numbers is: \begin{align*} \frac{1}{1}&+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\dotsb+\frac{1}{5050}\\ &=2\left[\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\dotsb+\frac{1}{10100}\right] \end{align*} The right hand side transforms as \begin{align*} 2&\left[\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+\dotsb+ \left(\frac{1}{100}-\frac{1}{101}\right)\right]\\ &=2\left(\frac{1}{1}-\frac{1}{101}\right) \end{align*} Now multiply by $3$ to get your desired answer: $6\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{600}{101}$

Daniel Buck
  • 3,564