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Geometric sequence question

Hey, just looking for a bit of help if you guys don't mind....
so a lad decides to invest 4k given from his rich conservative daddy into an account at the start of every year. Because of his daddy's connections and high prominence, the lad manages to get an interest rate of 1.04% which is added at the end of the year. how much is the total after 5 years?

This question got me confused because it says 4000 is added every year and on top of that there is an interest rate.
I know to sub this into the equation S(n) = a((r^n)-1)/r-1
However, I am very confused as to what to sub in, as 4000 is added every year. I was thinking the equation would change to be, S(n) = na(r^n-1)/r-1 but this is wrong. Can someone help me on what to sub in? I can find out the common ratio and the first term very easily but the +4k per year has puzzled me... thanks in advance.
Reply 1
Avatar for Notnek
Notnek
21
Original post by differentiation
Hey, just looking for a bit of help if you guys don't mind....
so a lad decides to invest 4k given from his rich conservative daddy into an account at the start of every year. Because of his daddy's connections and high prominence, the lad manages to get an interest rate of 1.04% which is added at the end of the year. how much is the total after 5 years?

This question got me confused because it says 4000 is added every year and on top of that there is an interest rate.
I know to sub this into the equation S(n) = a((r^n)-1)/r-1
However, I am very confused as to what to sub in, as 4000 is added every year. I was thinking the equation would change to be, S(n) = na(r^n-1)/r-1 but this is wrong. Can someone help me on what to sub in? I can find out the common ratio and the first term very easily but the +4k per year has puzzled me... thanks in advance.

E.g. after the third year, the first year's 4000 has gained interest for 3 years, the second year's 4000 has gained interest for 2 years and the third for 1 year.

Does that help? Also, do you have to use geometric sequences for this question? - it could be done by a GCSE student who has studied compound interest.
(edited 9 years ago)
Original post by differentiation
Hey, just looking for a bit of help if you guys don't mind....
so a lad decides to invest 4k given from his rich conservative daddy into an account at the start of every year. Because of his daddy's connections and high prominence, the lad manages to get an interest rate of 1.04% which is added at the end of the year. how much is the total after 5 years?

This question got me confused because it says 4000 is added every year and on top of that there is an interest rate.
I know to sub this into the equation S(n) = a((r^n)-1)/r-1
However, I am very confused as to what to sub in, as 4000 is added every year. I was thinking the equation would change to be, S(n) = na(r^n-1)/r-1 but this is wrong. Can someone help me on what to sub in? I can find out the common ratio and the first term very easily but the +4k per year has puzzled me... thanks in advance.


I have to find the sum of it.. I dont know how to put that into the equation for the sum of geometric sequences. I can't seem to make sense of it for some reason.
I think each term would go like (a) + (ar+a) + (r(ar+a)+a) + (r(r(ar+a)+a) + a)
Reply 4
Avatar for Notnek
Notnek
21
Original post by differentiation
I think each term would go like (a) + (ar+a) + (r(ar+a)+a) + (r(r(ar+a)+a) + a)

That's right and if you expand them you get

ar0+(ar0+ar1)+(ar0+ar1+ar2)+...ar^0+(ar^0+ar^1) + (ar^0 + ar^1+ar^2) + ...

Can you see how each part is the sum of a geometric sequence?
Reply 5
Avatar for ztibor
ztibor
10
Original post by differentiation
I have to find the sum of it.. I dont know how to put that into the equation for the sum of geometric sequences. I can't seem to make sense of it for some reason.


The sum at the end of the 1st year: 4000(1+0.0104)4000(1+0.0104)
The sum at the end of the 2nd year: (4000(1+0.0104)+4000)(1+0.0104)=(4000(1+0.0104)+4000)(1+0.0104)=
=4000(1+0.0104)2+4000(1+0.0104)=4000(1+0.0104)^2+4000(1+0.0104)
.....
At the end of the 5th year

4000(1+0.0104)5+4000)(1+0.0104)4+4000(1+0.0104)3+...+4000(1+0.0104)=4000(1+0.0104)^5+4000)(1+0.0104)^4+4000(1+0.0104)^3+...+4000(1+0.0104)=
4000⋅1.0104(1+1.0104+...+1.01044)=4000⋅1.01041.01044−10.01044000\cdot 1.0104(1+1.0104+...+1.0104^4)=4000\cdot 1.0104\frac{1.0104^4-1}{0.0104}

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