money question (1 Viewer)

[yellow_angel]

Initially Joan puts in $1000 in the bank.
The money is compounded by 1% daily, at the end of each day.
On the second day she puts in $999...
And she puts in $1 less everyday.

How much money does Joan have in her bank account on the thousandth day?
'leave it unsimplified' (Hint: Use integration)


Any ideas?

 

[Stefano]

Any ideas?

Yes - ask the bank.


(I hate these questions, sorry!)

 

[Vampire]

where'd u get that q? seems pretty damn hard...

 

[serge]

i think it goes like this

1,000(1.01)^1000 since it'll be there for 1,000 days

+ 999(1.01)^999 same reason +.... 1(1.01)

so that should add up to

[1,000 + 999 +... 1]{ (1.01)^1000 + (1.01)^999 +... 1.01)}

the first bracket is an arithmetic series a=1000 d= -1 n=1000

and the second bracket is a geometric series a=1.01 r=1.01 n=1000

use the Sum of formulas and multiply

(i think this will figure out what she has on the END of the 1,000th day
not sure if it'll make much difference)

 

[rama_v]

Initially Joan puts in $1000 in the bank.
The money is compounded by 1% daily, at the end of each day.
On the second day she puts in $999...
And she puts in $1 less everyday.

How much money does Joan have in her bank account on the thousandth day?
'leave it unsimplified' (Hint: Use integration)


Any ideas?

Lety A_n be the amount in the bank after n days

so A₁ = 1000(1.01)
A₂ = 1000(1.01)² + 999(1.01)
A₃ = 1000(1.01)³ + 999(1.01)² + 998(1.01)
.
.
.
A_n = 1000(1.01)ⁿ + 999(1.01)^n-1 + ... + 2(1.01)² + 1(1.01)

Now I dunno where to go

 

[menty]

Not sure if this is right

T1 = 1000(1.01)^1000
T2= 999(1.01)^999
T3=998(1.01)^998


blah


T1000 = 1(1.01)^1


Add all these up
S1000= (1.01)[1 + 1.01^2 + 1.01^3 + ... + 1.01^999][1+2+3+ .... + 1000]
= (1.01)(1.01^1000-1)/0.01 X 500X1001
= 1 X 10^12 loldollars

 

[serge]

can someone post the answer?

using my above method i got

500(1001).(1.01[1.01^1000 -1]/0.01)
which becomes roughly 1.059 X10^12

S1000= (1.01)[1 + 1.01^2 + 1.01^3 + ... + 1.01^999][1+2+3+ .... + 1000]
= (1.01)(1.01^1000-1)/0.01 X 500X1001
= 1 X 10^12 loldollars

cool, we both got the same thing, hope its right

 

[yellow_angel]

k... I sum how managed to find the answers brace ur selves


A1=1000 (think it's because it's end of the day compouded)

A1000=1000(1.01)^999+999(1.01)^998+...+1 (da final dollar)

let x=1.01

INT A1000 .dx = x^1000+x^999+x^998+...+x+c (constant)
= [x(x^1000-1)]/(x-1) + c (geomaterical series sum formula)

THEREFORE

A1000= Differentiation of "[x(x^1000-1)]/(x-1) + c" (constant now dissappears)
= (1000x^1001-1001x^1000+1)/[(x-1)^2] (using the quotient rule)

The money is (1000(1.01)^1001-1001(1.01)^1000+1)/[(0.01)^2]

holy ***

 

[robbo_145]

The money is (1000(1.01)^1001-1001(1.01)^1000+1)/[(0.01)^2]

holy ***

yeah my answer agrees with yours (though i didnt use integration)

1,000(1.01)^1000 since it'll be there for 1,000 days

+ 999(1.01)^999 same reason +.... 1(1.01)

so that should add up to

[1,000 + 999 +... 1]{ (1.01)^1000 + (1.01)^999 +... 1.01)}

that does not add to that
you have forgotton all the internal terms
(a+b+c)(x+y+z) =/= ax + by + ca

it = ax + ay + az + bx + by + bz + cx + cy + cz

 

[Ghost1788]

does she put the money in before or after the interest is added/calculated?