Complex summation

[chewrn]

Hi, I have this math problem that I've been thinking... and it keeps coming back to me, so I want to solve this problem.

Say I earn $300 on the first day. $700 on the second day, $1200 on the third day.
And this goes up until 50th day.
I want to know the total sum on the 50th day.

Let f(x) be the amount of money I earn, whereas x is the day.
Σf(x=1to50)= 300 + (400+f(1)) + (500+f(2))... (5200+f(49))
Σf(x=1to50)= 300_x1 + [((f(1)+100(x-1))+f(x-1))]_x2 + [((f(1)+100(x-1))+f(x-1))]_x3... +[((f(1)+100(x-1))+f(x-1))]_x49

This is what I have so far. I could program and calculate using recursion, but I want to know if there is a way to do this quickly by hand.

Any help to this problem will be greatly appreciated.

 

[chewrn]

300     700     1200      1800        2500 ......
    400     500      600        700 ......
        100      100       100 ....

300n + 400n(n-1)/2 + 100n(n-1)(n-2)/6

With your problem, n = 50.
Substitute and you'll get 2,465,000

If you need the amount contributed ON the 50th day:
300 + 400(n-1) + 100(n-1)(n-2)/2
Substitute n=50 and you'll get 137,500

Thank you for the reply. Could you show me the steps on how you ended up with those functions?

 

[HallsofIvy]

I can't be certain how Denis did it but here is how I would do it:
Taking "the first day" to be n= 1, the second day to be n= 2, etc. On "day 0" you earned $0.
The "total sum" on the zeroth day was f(0)= 0, first day is f(1)= 300, f(2)= 300+ 700= 1000, f(3)= 1000+ 1200= 2200.

The "first differences" are \(\displaystyle \Delta f(0)= 300- 0= 300\), \(\displaystyle \Delta f(1)= 1000- 300= 700\(\displaystyle and \(\displaystyle \Delta f(2)= 2200-1000= 1200\). Those are, obviously, the amounts earned the first two days.\).

The "second differences" are \(\displaystyle \Delta^2 f(0)\)= 700- 300= 400, \(\displaystyle \Delta^2(1)= 1200- 700= 500.

The "third differences" is \(\displaystyle \Delta^3 f(0)\)= 500- 400= 100.

That is as far as we can go but the wording certainly implies that "this goes on" so we can presume that the third difference is always 100 and, therefore, the fourth and higher differences are 0.

By "Newton's divided difference formula", the function is given by \(\displaystyle f(0)+ \Delta f(0) n+ (\Delta^2 f(0)/2)(n)(n-1)+ (\Delta^3 f(0))/6= 300+ 300n+ (400/2)n(n- 1)+ (100/6)n(n-1)(n-2)\)
\(\displaystyle = 300+ 300n+ 200(n^2- n)+ (50/3)(n^3- 3n^2+ 2n)= 300+ 400n+ (50/3)n^3- 50n^2+ (100/3)n= (50/3)n^3- 50n^3+ (1150/3)n+ 300\).

(Denis took n= 1 for the first month while I took n= 0. To find the value on the fiftieth day, set n= 49 in my formula.)\)\)

 

[chewrn]

I can't be certain how Denis did it but here is how I would do it:
Taking "the first day" to be n= 1, the second day to be n= 2, etc. On "day 0" you earned $0.
The "total sum" on the zeroth day was f(0)= 0, first day is f(1)= 300, f(2)= 300+ 700= 1000, f(3)= 1000+ 1200= 2200.

The "first differences" are \(\displaystyle \Delta f(0)= 300- 0= 300\), \(\displaystyle \Delta f(1)= 1000- 300= 700\(\displaystyle and \(\displaystyle \Delta f(2)= 2200-1000= 1200\). Those are, obviously, the amounts earned the first two days.\).

The "second differences" are \(\displaystyle \Delta^2 f(0)\)= 700- 300= 400, \(\displaystyle \Delta^2(1)= 1200- 700= 500.

The "third differences" is \(\displaystyle \Delta^3 f(0)\)= 500- 400= 100.

That is as far as we can go but the wording certainly implies that "this goes on" so we can presume that the third difference is always 100 and, therefore, the fourth and higher differences are 0.

By "Newton's divided difference formula", the function is given by \(\displaystyle f(0)+ \Delta f(0) n+ (\Delta^2 f(0)/2)(n)(n-1)+ (\Delta^3 f(0))/6= 300+ 300n+ (400/2)n(n- 1)+ (100/6)n(n-1)(n-2)\)
\(\displaystyle = 300+ 300n+ 200(n^2- n)+ (50/3)(n^3- 3n^2+ 2n)= 300+ 400n+ (50/3)n^3- 50n^2+ (100/3)n= (50/3)n^3- 50n^3+ (1150/3)n+ 300\).

(Denis took n= 1 for the first month while I took n= 0. To find the value on the fiftieth day, set n= 49 in my formula.)\)\)

\(\displaystyle \(\displaystyle

Thank you this clears it up for me \)\)