Chapter 7.1, Problem 64PS

Solution

To state: The money you will have saved after 100 days and the number of days you must save to have saved $ 500 if you want to save $500 for a school trip. You begin saving a penny on the first day. You plan to save an additional penny day after that.

For example, you will save 2 pennies on the second day, 3 pennies on the third day, and so on.

The resultant sum after 100 days is 5050 and $500 can be saved in approx 32 days.

Given information:

You want to save $ 500 for a school trip. You begin saving a penny on the first day. You plan to save an additional penny day after that.

Explanation:

You begin saving a penny on the first day. You plan to save an additional penny day after that.

Then, the saved amount after 100 days is:

  $1+2+3+\cdots +100={\displaystyle \sum _{i=1}^{100}i}$

Now find the sum,

  $\begin{array}{c}{\displaystyle \sum _{i=1}^{100}i}=\frac{100\left(100+1\right)}{2}\\ =\frac{100\cdot 101}{2}\\ =50\cdot 101\\ =5050\end{array}$

The amount saved after 100 days is 5050 pennies.

The number of days needed to save $500 is:

  $\begin{array}{c}{\displaystyle \sum _{i=1}^{n}i}=\frac{n\left(n+1\right)}{2}\\ 500=\frac{n^2+n}{2}\\ 1000=n^2+n\\ 0=n^2+n-1000\end{array}$

Now solve this quadratic equation,

  $n^2+n-1000=0$

  $\begin{array}{l}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a},a=1,b=1,c=-1000\\ x=\frac{-1\pm \sqrt{1^2-4\left(1\right)\left(-1000\right)}}{2}\\ =\frac{-1\pm \sqrt{1+4000}}{2}\\ =\frac{-1\pm \sqrt{4001}}{2}\end{array}$

The values are:

  $\begin{array}{l}x=\frac{-1-\sqrt{4001}}{2}\approx -32.1267\\ x=\frac{-1+\sqrt{4001}}{2}\approx 31.1267\end{array}$

The value can’t be considered in negative therefore the number of days are approx 32.