Question

Can someone clarify this question... is he investing the 10K as he gets it
You have a trust fund that will become available to you three years from now (at the end of the third year). The fund will pay $10,000 every six months starting at the end of the 3rd year for 10 years (to the end of the 13th year). If you invest all of this money into a savings account that has an interest rate of 8% compounded weekly how much money will you have 30 years from now?

Answer 1

..........

Answer 2

what -.-

Answer 3

its this random economics course theyre making us all take

Answer 4

dang been forever since I solved an economics problem.

Answer 5

dont need to solve it

Answer 6

just need to know what the question is asking, with the assumption i made it became this hugeeee problem

Answer 7

so im rethinking my assumption

Answer 8

for 10 years, u will be getting 10K 20 times. (twice a year)

Answer 9

you're investing it to a savings account that pays 8% interest compounded weekly

Answer 10

for example :-
ur first check : 10K at the start of 4th year yields
10K(1+.08/52)^[52*(30-3)]

in 30 years from now.

Answer 11

right so

Answer 12

form a series

Answer 13

everytime he gets the 10k is that added to what he is already investing

Answer 14

yes, every 6 months he adds 10K to his savings account

Answer 15

alright gotcha thanks

Answer 16

this is called annuity problem..
simplifies to a nice geometric series

Answer 17

ya i noticedd

Answer 18

i got a formula like this

Answer 19

kj + (kj^2+kj) + (kj^3+kj^2+kj)+...
first 2nd 3rd

Answer 20

*pretends to knows what everyone is say*

-nods-

Answer 21

in that form
where my j=1+(r/n)^(n*k/20)

Answer 22

does that look right i mean the series

Answer 23

saying* smh

Answer 24

and i did some random math to try and solve for that series

Answer 25

so there should be
20kj + 19kj^2+18kj^3.... like that, and i solved for a formula that gives me the sum of this

Answer 26

but it was a lot of math, i might have made many random mistakes here and there xD

Answer 27

4_a : 10K(1+.08/52)^[52*(30-3)]
4_b : 10K(1+.08/52)^[52*(30-3-1/2)]

5_a : 10K(1+.08/52)^[52*(30-3-2/2)]
5_b : 10K(1+.08/52)^[52*(30-3-3/2)]
...

Answer 28

we need to add them all.
does that look right

Answer 29

waiitt how come u can do that

Answer 30

it simplifies to ur form, im sure..

Answer 31

interestingg i was thinking about this like...

Answer 32

4_a :
im getting 10K in 3years from now.
and ima invest it in bank for (30-3) years hmm...

Answer 33

are u allowed to think about it as as
10k being compounded n times + 10K again being compounded n-1 times +10K...

Answer 34

yes we need to think like that oly. cuz we are getting money every 6 months.

Answer 35

ohh realllyyy

Answer 36

oncee second!! give me 2 mins i need to think about that

Answer 37

ok.. even i need to think about ur equations.. :)

Answer 38

i dont why but i just kept thinking that cannot be possible.. with the whole compouding thing,, doesnt compounding 20K at once differ from compounding 10k and 10k

Answer 39

how ?

20K (1 + r/n)^t = 10K(1+r/n)^t + 10K(1+r/n)^t

Answer 40

oh ya that is true but wait thats not what i meant

Answer 41

for the 2nd compound it would be a function of the compound before

Answer 42

so i was thinking how you can write it as separate compounding

Answer 43

these business problems are nastay :P

Answer 44

because by the time he is trying to find the compound for the 2nd term, there is some extra money that is being compounded that he isnt concerning himself with right?? or is he calculating that part too with the inital formula

Answer 45

he is calculating that part too.
think of it like this : each 10K packet has its own life. compounds separately itself.
you dont need to add every 6 months all ur investments. it all turns out to be same...

Answer 46

4_a : 10K(1+.08/52)^[52*(30-3)]
4_b : 10K(1+.08/52)^[52*(30-3-1/2)]

5_a : 10K(1+.08/52)^[52*(30-3-2/2)]
5_b : 10K(1+.08/52)^[52*(30-3-3/2)]
...

Answer 47

4_a : 10K(1+.08/52)^[52*(30-3)]
is taking into account compounding effects also for first 10K

Answer 48

ya you are right arrghh

Answer 49

mann i wasted so much time not thinking about this part lol

Answer 50

i did all this crazy math avoiding that simplification

Answer 51

basically, you can pull out 10K, and wat remains is a simple geometric series..

Answer 52

lol... i spent more than a week on this a year ago.... to make sense of annuity setup... :o

Answer 53

look at my picture!! its crazy lol i found this really intersting way though

Answer 54

to deal with this
sum
20kj+19kj^2+18kj^3....

Answer 55

that will probably come in handy... ahemmm

Answer 56

im still trying to understand.. lol its mouthful

Answer 57

20kj+19kj^2+18kj^3.... +and- (kj^2+2kj^3....)

Answer 58

so thank i can write it as
20k(j+j^2+j^3...) - k(j^2+2j^3+3j^4....)
then repeat the same process on the right side again

Answer 59

then there was this series pattern that emerged out of this lol
for the first time i saw the whole + and - alternating series pop out randomly

Answer 60

i found the formula for sum of odd geometric series didnt even think it was possible till i realized this cool simplifcation to be made

Answer 61

k(j^3+j^5+j^7...)
=kj(j^2+j^4+j^6...)
=kj((j^2)^1+(j^2)^2+(j^2)^3...) = kj(j^2-(J^2)^n/1-j^2)

Answer 62

this is all probably really noob to you though lol

Answer 63

i just found it really fun... to do this question.. i felt like oh wow finally i am using all that random knowledge about gemoetric series haha

Answer 64

oh you're adding up previous accumulated money to 10K each time, and simplifying it... looks nice :) it comes very handy when u do a problem like below :-

Suppose you have taken 500K for loan, and you want to pay back in 10 years at an interest rate of 10% compounded quarterly. Calculate the EMI.

Answer 65

im going for lunch... cya :)

Answer 66

cya! sleep time for me

Answer 67

gn have good sleep :D

Answer 68

ohh this is another way... the geometric series formulas will simplify in the end
20k(j+j^2+j^3...) - k(j^2+2j^3+3j^4....)
=20k(j+j^2+j^3...) - k(j^2+j^3+j^4....)-k(j^3+j^4+...)-k(j^4+j^5+...)-k(......
=20k(j-j^21/1-j) - k (J^2-j^21/1-j)-k(j^3-j^21/1-j)....
=20k(j-j^21/1-j) - [k/(1-j)] *-1*(2+3+4...+20) + 20*(k/1-j)*j^21
=20k([(j-j^21)/(1-j)]- [k/(1-j)]*-1*((22*19)/2) + 20*(k/(1-j))*j^21

Answer 69

well this stuff is pretty much impossible to see without latex.. or drawing