# Infinite geometric series

1. Nov 26, 2015

### Pual Black

1. The problem statement, all variables and given/known data
hello this question is discussed in 2009 but it is closed now

If you invest £1000 on the first day of each year, and interest is paid at 5% on
your balance at the end of each year, how much money do you have after 25
years?

2. Relevant equations
$S_N=\sum_{n=0}^{N-1} Ar^n$

where N is the last term
r is the common ratio & A is a constant

$S_N= a\frac{1-r^N}{1-r}$

3. The attempt at a solution

after 25 years i would set N=25 but this will give me a result of £47727. then i have to subtract £1000 because on the first day of each year i invest £1000 therefore i got a result of £46727

but this is the wrong answer

if i set N=26 i will get £51113 and then again subtract £1000
therefore i got a result of £50113 and this is the right answer

my question is why i must put N=26. isn't N the last term and equal to 25?
and is this way of solution right?

2. Nov 26, 2015

### BvU

Can't follow your equations. "N is the last term"? Wouldn't N be the number of terms/years ?
And then after one year you don't have $Ar^0$ but $Ar^1$.

Tip: Don't change the notation from one line to the next. A is A, not a.

$S_N= A\frac{1-r^N}{1-r}$ looks weird too. Don't you mean $S_N= Ar\frac{\ 1+r^N}{1+r}$ so that after 1 year you do have Ar ?

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3. Nov 26, 2015

### Ray Vickson

No, his formula is correct as written: $\sum_{n=0}^N A r^n = A(r^N -1)/(r-1)$ is a standard elementary algebra result. Of course, it also equals $A(1-r^N)/(1-r)$.

4. Nov 26, 2015

### ehild

If you sum from 0 to N, it is N+1 elements. The correct formula is $\sum_{n=0}^{N-1} A r^n = A(r^N -1)/(r-1)$

Last edited: Nov 27, 2015
5. Nov 27, 2015

### Ray Vickson

Indeed: that was a typo on my part. I had intended to make the upper summation limit equal N-1, but somehow slipped up.

6. Nov 27, 2015

### BvU

Sorry for the minus sign - weak moment.

Time to define what A stands for. My impression was that after one year the guy has $Ar$, not $\sum_{n=0}^{0} A r^n = A(r^1 -1)/(r-1) = A$

Time to define what N stands for, too: at the begining of year 2, they guy has
$\sum_{n=0}^{1} A r^n = A(r^2 -1)/(r-1) = A(r+1)$ and he invests another A ?

I figured r = 1.05 and A = 1000 pounds. Where did I go all wrong ?

$$1000 \,{1.05^{26} - 1\over 1.05 -1} = 51113.45$$
$$1050 \,{1.05^{25} - 1\over 1.05 -1} = 50113.45$$ which OP considered the right answer .

The
in the template is clearly useful. And includes the list of what values in the problem statement the symbols used stand for ...

Last edited: Nov 27, 2015
7. Nov 27, 2015

### Pual Black

yes you are right N is the number of terms
and i get it now. first i invest £1000 so i have nothing at this moment after one year i have £1000*1.05= £1050
Easier: just write it out correctly from the start. If $r = 1.05$, $A = 1000$ and $N = 25$, the value you want is the "future value" $F$:
$$F = A r + A r^2 + \cdots + A r^N = A r \sum_{n=0}^{N-1} r^n$$
Basically, the final $A$ at the start of year 25 becomes $Ar$ at the end of year 25. The second-last $A$ at the start of year 24 becomes $A r^2$ at the end of year 25, and so forth.