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Hey people, i am trying to figure out a formula to calculate something, which is the following:
If i put 600 euros a year on a bank account, i get a 5% interest a year over that.
that would be 600*1,05 for the first year, ((600*1,50)+600)*1,05 for the second year, but how do i calculate that with a formula, an easy way over lots of years? simpely using ^30 for 30 years for example.

thnx in advance ppl ^^

(600x1.05)^30 ,, thats it

NOTE -- I don't really undertsnad the question at all, here is my interpretation. aka I do not know what I am going on about.

Hmmm...I don't even know why I'm thinking Maths is one of my worst subjects.

But from what I'm hearing you're trying to find a formulea that shows how much interest you get back over a period of time with the amount a 600 and the rate at 5%.

Well first, 5% of 600 is 30, that was easily found by 600/100*5

So a formula could be (X/100*Y)+X=B

Letting X be the amount, Y then interest rate and B the total for that year.

Multiple years could be -

A[(X/100*Y)+X]=B

Let A be the number of years.

This is probably all incorrect, but It could give you some ideas, sorry if I could not help.

Math to me, is like -

whiteraven wrote:(600x1.05)^30 ,, thats it

nope, that is like 600*1,05 =630 and again = 1260 instead of 1291,50(What it should be) o.o IF it would have been times btw, your (600*1,05)^30 = 9,55469E+83 , aint really right =O

CloudStrider wrote:NOTE -- I don't really undertsnad the question at all, here is my interpretation. aka I do not know what I am going on about.

Hmmm...I don't even know why I'm thinking Maths is one of my worst subjects.

But from what I'm hearing you're trying to find a formulea that shows how much interest you get back over a period of time with the amount a 600 and the rate at 5%.

Well first, 5% of 600 is 30, that was easily found by 600/100*5

So a formula could be (X/100*Y)+X=B

Letting X be the amount, Y then interest rate and B the total for that year.

Multiple years could be -

A[(X/100*Y)+X]=B

Let A be the number of years.

This is probably all incorrect, but It could give you some ideas, sorry if I could not help.

Math to me, is like -

A[(X/100*Y)+X]=B

A the number of years.
B Total
X amount
Y interest

this would add the last "X" or amount or 600 euros without taking interest over it, so that aint right either =O thnx for the feedback anyhow though

Shit you're right. -.-' </3

CloudStrider wrote:Shit you're right. -.-' </3

ahh ty for trying anyway xD if ppl dont try ill never get my awnser xD even false awnsers with good logics behind it might help ^^

X.Y^A=B

==>600.1,05^1=630 for 1 year
==>600.1,05^2=661,5 for 2 years
==>600.1,05^3=694,575 for 3 years
etc


A The number of years
B Total
X Amount
Y Interest

Stefaab wrote:X.Y^A=B

==>600.1,05^1=630 for 1 year
==>600.1,05^2=661,5 for 2 years
==>600.1,05^3=694,575 for 3 years
etc


A The number of years
B Total
X Amount
Y Interest

but, im putting in an additional 600 euros every year, which will also take interest. as explained in my first post that would be:
600*1,05 for the first year.
((600*1,50)+600)*1,05 for the second year.
but try to do that over 33 years -_-'

LillDev!l wrote:Hey people, i am trying to figure out a formula to calculate something, which is the following:
If i put 600 euros a year on a bank account, i get a 5% interest a year over that.
that would be 600*1,05 for the first year, ((600*1,50)+600)*1,05 for the second year, but how do i calculate that with a formula, an easy way over lots of years? simpely using ^30 for 30 years for example.

thnx in advance ppl ^^

you should try:

(600x1.05^33) + (600x1.05^32) + (600x1.05^31) etc.

xD goodluck xD now, dont be lazy and go calculate it xD

whiteraven wrote:

LillDev!l wrote:Hey people, i am trying to figure out a formula to calculate something, which is the following:
If i put 600 euros a year on a bank account, i get a 5% interest a year over that.
that would be 600*1,05 for the first year, ((600*1,50)+600)*1,05 for the second year, but how do i calculate that with a formula, an easy way over lots of years? simpely using ^30 for 30 years for example.

thnx in advance ppl ^^

you should try:

(600x1.05^33) + (600x1.05^32) + (600x1.05^31) etc.

xD goodluck xD now, dont be lazy and go calculate it xD

600*1,05^33 = interest off 600 euros over 33 years (=3001,913125)
im looking for interst of 600 euros off the first year, + an additional 600 euros for the second year with interest etc.
so for 4 years = ((((((600*1,05)+600)*1,05)+600)*1,05)+600)*1,05=2715,37875

in this case, my 4 years would be almost as much as your 33 years of interst only.
but, adding two "("s for every year ill use like 66 off em xD, making the chance for little errors big aswell, leaves alone if i want to see after how it is if i put in 700 euros a year, instead of 600 euros =]

staat wel in je m&o boek

Jantje wrote:staat wel in je m&o boek

+ heb je al in de 1e gehad met wiskunde -.-

heroo wrote:

Jantje wrote:staat wel in je m&o boek

+ heb je al in de 1e gehad met wiskunde -.-

leg maar uit dan hé?:P (explain it then ^^)

ok so, after 33 years it would be :
((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((600*1,05)+600)*1,05)+600)*1,05)+600)*1,05)+600)*1,05)+600)*1,05)+600)*1,05)+600)*1,05)+600)*1,05)+600)*1,05)+600)*1,05)+600)*1,05)+600)*1,05)+600)*1,05)+600)*1,05)+600)*1,05)+600)*1,05)+600)*1,05)+600)*1,05)+600)*1,05)+600)*1,05)+600)*1,05)+600)*1,05)+600)*1,05)+600)*1,05)+600)*1,05)+600)*1,05)+600)*1,05)+600)*1,05)+600)*1,05)+600)*1,05)+600)*1,05)+600)*1,05= 50440,17563

-_-' ya, i know, i was bored

f(x)=600x1.05^x

Dat is de standaardformule, met x in aantal jaar.

Laat me ff denken over die extra 600 elk jaar =p

Volgens mij word het dan gewoon:
f(x)=(600x1.05+600)^x

Meh...Dat klopt volgens mij niet >.> Maar het is ook zaterdag dus

[SW]XMoshe wrote:f(x)=600x1.05^x

Dat is de standaardformule, met x in aantal jaar.

Laat me ff denken over die extra 600 elk jaar =p

Volgens mij word het dan gewoon:
f(x)=(600x1.05+600)^x

Meh...Dat klopt volgens mij niet >.> Maar het is ook zaterdag dus

het is ook een liniare formule en een expontentiële formule in 1 xD dus weet niet eens of het wel mogelijk is in 1 formule

LillDev!l wrote:

[SW]XMoshe wrote:f(x)=600x1.05^x

Dat is de standaardformule, met x in aantal jaar.

Laat me ff denken over die extra 600 elk jaar =p

Volgens mij word het dan gewoon:
f(x)=(600x1.05+600)^x

Meh...Dat klopt volgens mij niet >.> Maar het is ook zaterdag dus

het is ook een liniare formule en een expontentiële formule in 1 xD dus weet niet eens of het wel mogelijk is in 1 formule

Misschien f(x)=((600+(600x x))x1.05)^x? O_o (die x x is beetje vaag maargoed)

/Edit: Was haakjes vergeten.

Niet NL gaat praten, snappen velen niets van

lildevil vertel je opdracht nog eens in nederlands dan,, snap em niet helemaal

whiteraven wrote:lildevil vertel je opdracht nog eens in nederlands dan,, snap em niet helemaal

Hij wil een formule opstellen waarbij hij in het 1e jaar 600 euro op de bank zet tegen 5% rente, en elk jaar doet hij er 600 euro bij.

[SW]XMoshe wrote:

whiteraven wrote:lildevil vertel je opdracht nog eens in nederlands dan,, snap em niet helemaal

Hij wil een formule opstellen waarbij hij in het 1e jaar 600 euro op de bank zet tegen 5% rente, en elk jaar doet hij er 600 euro bij.

precies, waar ik dus ook weer rente over krijg =]

um getting back to english?:

its really just 600 * 1.05^30 years that way you have 600 * 4.32 = $2593.17 after 30 years meaning your money more than quadripled in 3 decades. yay for exponential functions

well actually thats if you only put in money once

this is if you put in money each year:

total = (600*1.05^30) + (600*1.05^29) + (600*1.05^28) +...+ (600*1.05^1)

total * 1.05 = (600*1.05^31) + (600*1.05^30) + (600*1.05^29) +...+ (600*1.05^2)

therefore:

total - (total*1.05) = (600*1.05^31)-(600*1.05^1) <since all the values in between cancel out>

total * (1 - 1.05) = (600*1.05^31)-(600*1.05^1)

total = ((600*1.05^31)-(600*1.05^1)) / (1 - 1.05)

total is the total amount of money you have after 30 years, 1.05 is the rate of growth

so the formula is basically:

total = ((600 * R^(x+1))-(600 * R^1)) / (1 - R)

each year +600 and +5% interest
=> arithmetico-geometric serie type u(n+1) = an + b

1st: Y1 = (0+600)*1.05 = 630
2nd: Y2 = (630+600)*1.05 = 1291.5
3rd: Y3 = (1291.5+600)*1.05 = 1986.075
etc...

Developping term relative to Y1:
Y2 = ((0+600)*1.05+600)*1.05 = 630 + 630*1.05
Y3 = (((0+600)*1.05+600)*1.05+600)*1.05 = 630 + 630*1.05 + 630*1.05²

=> year N, by conjecturation
Nth: YN = 630 + 630*1.05 + 630*1.05² + ... + 630*1.05^n

It's the sum of the geometric series
=> formula year n : 630*(1-1.05^n)/(1-1.05)

P.S: This is not the real demonstration (dont apply this in exam)

informatic ftw!