Without a lot of calculations, I would choose option 3 because I know that on the last day of the month I will have anywhere from 2^27 p to 2^30p (because I start from 2^0) depending on the month, which is far larger than £10 (or 1000 p) or 0.5(31)+2.5 = £18 that I would get on the last day with the other options.
So, to see how much would I have at the end of the month, I should find a series of how much I would get and find the value of it.
Firstly, option 1 gives me £10 per day. By the end of the 31 days, I would have 10*31= £310
In option 2, I would get an extra 50p (£0.5) each day, so £3 on day 1, £3.50 on day 2, £4 on day 3 and so on, thus being an arithmetic series. So the series is:
∑0.5x+2.5 between 1≤x≤31.
= (0.5∑x) + ∑2.5
= (0.5(31)(32)/2) + 31*2.5
(Formula for ∑x to a value of n = n(n+1)/2
= 248 + 77.5
= £325.50
So it seems that option 2 would be more beneficial to me in the span of a month.
However, option 3 would be the most beneficial for me by a mile. As the amount I get doubles each day, my series would be a geometric one: ∑2^(x-1) between 1≤x≤31
The formula for a finite geometric sequence is a(r^n-1)/r-1 where a is the first term of the series, r is the common ratio, and n is the final x value of the series.
In this case, a=1 since we start with 1p, r is 2 since the amount I get doubles every day, and n is 31, the number of days in the month.
Therefore, the series adds up to:
1(2^(31)-1)/2-1
= 2^(31)-1
= 2147483647p
≈ 2.1*10^9 p= £2.1*10^7.
So clearly, choosing option 3 will make me a millionaire in a month, therefore I will happily go for that.
Answers to the next few questions:
1) In which months would option 1 be better than option 2?:
Ans) Months will have 28, 29, 30 or 31 days
In February (non leap year, 28 days):
Option 1 would give me £10*28 =£280
Option 2 would give me 0.5(28)(29)/2+2.5(28)=£273
Therefore option 1 is better in a non leap year February by £7.
In February (leap year, 29 days):
Option 1 would give me £10*29 =£290
Option 2 would give me 0.5(29)(30)/2+2.5(29)=£290
Therefore, both the options will give me the same amount in a leap year February.
In a 30 day month:
Option 1 would give me £10*30 =£300
Option 2 would give me 0.5(30)(31)/2+2.5(30)=£307.5
Therefore, option 2 is better in a 30 day month by £7.5
In a 31 day month:
Option 1 would give me £10*31 =£310
Option 2 would give me 0.5(31)(32)/2+2.5(31)=£325.5
Therefore, option 2 is better in a 31 day month by £15.5
Over a 12 month period, with 4-30 day months, 7-31 day months and the 28 days of February, option 2 would be better since I would get 4(307.5)+7(325.5)+273= £3781.5 (Or £3798.5 on a leap year), whereas option 1 would only give me 4(300)+7(310)+280 = £3650 (or £3660 on a leap year)
2) If your family stopped your pocket money on day 8, which option would give you the most?:
Option 1 would give me 8*£10 = £80
Option 2 would give me 0.5(8)(9)/2 +2.5(8)= £38
Option 3 would give me (2^(8)-1) = 255p =£25.5
Option 1 would be a lot better in a span of 8 days.
3) On which day of the month does option 3 become the most fruitful?
The answer to this depends on what do we mean by "fruitful", do we mean when onwards will option 3 give me the most money on that particular day, or do we mean which day will our overall earnings of option 3 exceed the other options? Either way, we will have to solve 2 inequalities, they will just be different ones. The purpose of one will be to find when will option 3's earnings exceed option 1's and the other will be to find when will it exceed option 2's.
If the former is the intended question, the following will be the inequalities:
1) 2^(x-1)>10, (Option 3 vs. 1)
2) 2^(x-1)>0.5x+2.5, (Option 3 vs. 2)
I was unable to solve these inequalities algebraically, so I used desmos.com to answer them graphically.
The red curve is y=2^(x-1)
The blue line is y=10
And the green line is y=0.5x+2.5
As we can see, the red curve crosses the green line at x=3, which means that after the 4th day option 3 will be more fruitful per day than option 2.
Also, the red curve crosses the blue line at x=4.322, which means that on the 5th day, option 3 will become more fruitful per day than option 1, becoming the most fruitful option per day after that.
If the point of the question was to find the day when the overall earnings from option 3 would exceed the other 2 options, the inequalities would be as follows:
1) 2^(x)-1>10x, (Option 3 vs. 1)
2) 2^(x)-1>0.5(x)(x+1)/2 +2.5x, (Option 3 vs. 2)
I was unable to solve these algebraically either, therefore I used desmos.com again to solve these graphically
In the second graph:
The orange curve is y=2^(x)-1,
The black curve is y=10x
And, the blue curve is y=0.5(x)(x+1)/2 +2.5x
As shown in the graph, the orange curve crosses the blue curve first at x=3.648, so option 3 will become better than option 2 on the 4th day.
It also crosses the black line at x=5.909, which means that it will become a better option that option 1 on the 6th day, becoming the best option overall after that.
4) If you chose option 3, how many days would it be before you became a millionaire?:
£1,000,000 = 100,000,000p
Therefore, ∑2^(x-1) > 100000000
(When the summation happens between the interval 1≤x≤n, n being the day the sum satisfies the inequality)
Or, 2^(x)-1 > 10000000
2^(x) > 100000001
log(base 2)100000001 > x
Therefore, 26.5754
This means that I would become a millionaire on the 27th day.