You have £1,000 in your savings account. At the end of each month, the bank adds an extra 0.2% of interest to the outstanding balance and you make a withdrawal of £10.
How much is in the savings account after this has happened for 36 months, rounded to the nearest pound?
(I guess you need to solve it with a linear difference equation, but I cannot figure it out)

- anonymous

- jamiebookeater

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- anonymous

\[a ^{n}x _{0} +b (a ^{n}-1)/(a-1)\]
I found this formula for it and it gives the good answer but I do not see why.
(a^n=1.002^36 , xo=1000 , b=-10)

- anonymous

yeah i have to write this down. certainly cannot do it on the fly. if at all!

- anonymous

wait, why isn't it just \[1000(1.002)^{10}-10(1.002)^9\]?

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## More answers

- anonymous

actually my exponents are ridiculous. it is 36 months not ten. so i think maybe
\[1000(1.002)^{36}-10(1.002)^{35}\]

- anonymous

i just wrote out the first three months and saw what i got.

- anonymous

I am not sure here at all, I found that formula and it did give me the correct answer, but I can only see the answer unfortunately

- anonymous

first month P(1.002)
second month (P(1.002)-10)(1.002)=P(1.002)^2-10(1.002)
third month
(P(1.002)^2-10(1.002))-10)(1.002)=P(1.002)^3-10(1.002)^2-10(1.002)

- anonymous

oh wait maybe you get a geometric series for the ten part.

- anonymous

yeah that formula does look like a geometric series

- anonymous

first part is definitely p(1.002)^36

- anonymous

lets see what the last one is.

- anonymous

I think I see now why it is so

- anonymous

guess it depends on whether you take out ten pounds at the end. if so you will get 10+10(1.002)+10(1.002)^2+...+10(1.002)^35

- anonymous

\[x _{n}=a ^{n} x _{0} +(a ^{n-1} +....+a+1)b\]

- anonymous

and the last part is a series

- anonymous

equal to a^n -1/a-1

- anonymous

yeah i think so. i bet this simplifies to get essentially what i wrote at the beginning. if my algebra was good enuf i could do it

- anonymous

did you try 1000(1.002)^36-10(1.002)^35?

- anonymous

no, but I can

- anonymous

that gives a positive amount, that cannot be right, the interest is around 2 and you take out 10

- anonymous

we can use 1+r+r^2+...+r^35=\[\frac{r^{36}-1}{1-r}\] with r = 1.002

- anonymous

3.7289 rounded

- anonymous

yes that is what I thought too, but isnt the denominator r-1 ?

- anonymous

multiply by ten to get 37.289

- anonymous

yea typo

- anonymous

denominator in this case is .002

- anonymous

oh wait my calculation is wrong

- anonymous

i divided by .02

- anonymous

the answer was something around 700

- anonymous

i am off by a decimal

- anonymous

:) happens

- anonymous

\[\frac{(1.002)^{36}-1}{.002}=37.289\]

- anonymous

according to my calculator.

- anonymous

multiply by ten to get 372.89

- anonymous

according to my brain

- anonymous

subtract from 1000(1.002)^36 = 1074.578 rounded

- anonymous

gives the correct answer! sweet

- anonymous

and i get as a final answer (as they say) 701.68

- anonymous

whew.

- anonymous

glad we got this worked out even if a couple of false starts

- anonymous

maths is like this, try hard and fail a lot

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