Question

Geometric Sequences and Series:

Ok so I did a few of them and I am stuck in 1 question ... here sre my answers for the one I did.
A.)Please watch the following link, up until time 2:40. https://safeshare.tv/x/ss5a2ce1a1ef86a Use the appropriate formula to show how much you are paid on the 31st day if you choose the penny a day option.
My answer: 10737418.24
B. Use the appropriate formula to show your total earnings for the month if you choose the penny a day option
My answer: 21474836.48

D.) This question is the one I have trouble on...
Suppose you had this job for an infinte amount of time. Can you calculate your total earnigs if you are paid with the penny a day option. Why or why not.


This is how far I got.. a formula I think should be used is s=a1/1-r

R=2
A1=-0.01
N=31

PLEASE HELP! It doesnt have to be a full answer just give me tips on how to solve it if you can. Thank you :)

Answer 1

The video link doesn't work for me, but I assume this is the problem involving choosing between two payment types, one that immediately grants you some lump sum of cash, while the other starts off with $0.01 on the first day, $0.02 on the second day, $0.04 on the third, and so on, doubling per day.
Let
`m_n` be the amount of money earned on the
`n`-th day. Then
`m_n` is a geometric sequence that satisfies


`\begin{cases}m_1=0.01\\m_n=2m_(n-1)&\text{for }n>1\end{cases}`

We can solve explicitly for
`m_n` in terms of the starting payment
`m_1`:


`m_n=2m_(n-1)`

`m_(n-1)=2m_(n-2)\implies m_n=2^2m_(n-2)`

`m_(n-2)=2m_(n-3)\implies m_n=2^3m_(n-3)`

`\cdots`

`m_2=2m_1\implies m_n=2^(n-1)m_1`

So the amount of money earned on the
`n`-th day is


`m_n=2^(n-1)\cdot0.01`

Denote the total amount of money earned over
`n` days by
`M_n`. Then we can write


`M_n=\displaystyle\sum_(i=1)^nm_i=m_1+m_2+\cdots+m_(n-1)+m_n`

but since we have equivalent expressions for
`m_i` on any given day
`i`, this is the same as


`M_n=\displaystyle\sum_(i=1)^nm_i=0.01+2\cdot0.01+\cdots+2^(n-2)\cdot0.01+2^(n-1)\cdot0.01`

`M_n=0.01(1+2+\cdots+2^(n-2)+2^(n-1))`

Let's call the sum on the right hand side
`S_n`. Notice that

`S_n=1+2+\cdots+2^(n-2)+2^(n-1)`


`2S_n=2+2^2+\cdots+2^(n-1)+2^n`

`\implies S_n-2S_n=1-2^n`

`\implies-S_n=1-2^n`

`\implies S_n=2^n-1`

which means we end up with


`M_n=0.01(2^n-1)`


To answer part (D), you need to look no further than the formula for
`M_n`. The question is basically asking what happens as
`n` gets arbitrarily large. It should be clear that
`2^n` grows without bound, so as
`n\to\infty`, the amount of money you would get would diverge to infinity. So technically, you cannot calculate the amount because (1) there's only a finite amount of money to go around, and (2) infinity is not a "computable" number.
If, however, the scale factor used on your income was smaller than 1 - for example, say you were started with a million dollars on the first day, then your income got halved each day - then
`m_n` would eventually converge to 0, and on top of that,
`M_n` would converge to a finite number.