anonymous
  • anonymous
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)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • 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
  • anonymous
yeah i have to write this down. certainly cannot do it on the fly. if at all!
anonymous
  • anonymous
wait, why isn't it just \[1000(1.002)^{10}-10(1.002)^9\]?

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anonymous
  • 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
  • anonymous
i just wrote out the first three months and saw what i got.
anonymous
  • 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
  • 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
  • anonymous
oh wait maybe you get a geometric series for the ten part.
anonymous
  • anonymous
yeah that formula does look like a geometric series
anonymous
  • anonymous
first part is definitely p(1.002)^36
anonymous
  • anonymous
lets see what the last one is.
anonymous
  • anonymous
I think I see now why it is so
anonymous
  • 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
  • anonymous
\[x _{n}=a ^{n} x _{0} +(a ^{n-1} +....+a+1)b\]
anonymous
  • anonymous
and the last part is a series
anonymous
  • anonymous
equal to a^n -1/a-1
anonymous
  • 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
  • anonymous
did you try 1000(1.002)^36-10(1.002)^35?
anonymous
  • anonymous
no, but I can
anonymous
  • anonymous
that gives a positive amount, that cannot be right, the interest is around 2 and you take out 10
anonymous
  • anonymous
we can use 1+r+r^2+...+r^35=\[\frac{r^{36}-1}{1-r}\] with r = 1.002
anonymous
  • anonymous
3.7289 rounded
anonymous
  • anonymous
yes that is what I thought too, but isnt the denominator r-1 ?
anonymous
  • anonymous
multiply by ten to get 37.289
anonymous
  • anonymous
yea typo
anonymous
  • anonymous
denominator in this case is .002
anonymous
  • anonymous
oh wait my calculation is wrong
anonymous
  • anonymous
i divided by .02
anonymous
  • anonymous
the answer was something around 700
anonymous
  • anonymous
i am off by a decimal
anonymous
  • anonymous
:) happens
anonymous
  • anonymous
\[\frac{(1.002)^{36}-1}{.002}=37.289\]
anonymous
  • anonymous
according to my calculator.
anonymous
  • anonymous
multiply by ten to get 372.89
anonymous
  • anonymous
according to my brain
anonymous
  • anonymous
subtract from 1000(1.002)^36 = 1074.578 rounded
anonymous
  • anonymous
gives the correct answer! sweet
anonymous
  • anonymous
and i get as a final answer (as they say) 701.68
anonymous
  • anonymous
whew.
anonymous
  • anonymous
glad we got this worked out even if a couple of false starts
anonymous
  • anonymous
maths is like this, try hard and fail a lot

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