Question

You push your brother on a rope swing once over a creek and then let him swing freely. On the firstswing your brother travels a distance of 23 feet. On each successive swing, he travels 75% the distanceof the previous swing. What is the total distance your brother swings?

Answer 1

Let D(n) be the distance traveled on the nth swing. Then we have:

$\begin{gathered} D(1)=23\text{ ft} \\ D(2)=D(1)\cdot0.75\text{ ft} \\ D(3)=D(2)\cdot0.75^2\text{ ft} \\ D(n)=D(n-1)\cdot0.75^{n-1}\text{ ft} \end{gathered}$

This is a geometric progression. Then, if N is the total number of swings, the total distance is given by:

$\begin{gathered} \sum_{i\mathop{=}1}^ND(i)=\frac{D(1)\cdot(0.75^N-1)}{0.75-1}=\frac{23\cdot(0.75^N-1)}{0.75-1} \\ \therefore\lim_{N\to\infty}\frac{23(0.75^{N}-1)}{0.75-1}=\frac{23}{0.25}=92\text{ ft} \end{gathered}$