Thread: geometric progression

1. geometric progression

Hello
Can someone explain why multiplying the second formula by 'r' in the derivation of this formula doesn't change things in any way before subtracting?
...or just some background on the formula
Thanks

2. Originally Posted by 200001
Hello
Can someone explain why multiplying the second formula by 'r' in the derivation of this formula doesn't change things in any way before subtracting?
...or just some background on the formula
Thanks
It does change things slightly. You had something like $1+r+r^2+r^3+r^4+....+r^n$

when you multiply by r you get $r+r^2+r^3+r^4+...+r^n+r^{n+1}$

see how the we no longer have the 1? but we picked up that last term, then you subtract

3. proof of sum of geometric series

I guess I should make this formal to explain it better

If we are dealing with finite series we have this

$\displaystyle\sum_{k=0}^n ar^k=s=a+ar+ar^2+...+ar^n$

So that means $sr=ar+ar^2+...+ar^{n+1}$

So now lets subtract the two

$s-sr=(a+ar+ar^2+...+ar^n)-(ar+ar^2+...+ar^{n+1})$

$=a-ar^{n+1}=a(1-r^{n+1})$

So $s-sr=s(1-r)=a(1-r^{n+1})$

And so $s=\frac{a(1-r^{n+1})}{1-r}$

So multiplying by $r$ did change something, we didnt have the $1$ and we picked up a $r^{n+1}$

Now what if the series is infinite? Just take the limit as n goes to infinity. If $|r|<1$ then the $r^{n+1}$ tends to zero and so the formula becomes $\frac{a}{1-r}$

4. Thanks very much for taking the time to write that. It have been a great help

Now, if your series have started at 1 instead of 0 would that then leave use with R^n in the final summation in the numerator instead of the R^n+1 ?

Again, thanks for this

5. Originally Posted by 200001
Thanks very much for taking the time to write that. It have been a great help

Now, if your series have started at 1 instead of 0 would that then leave use with R^n in the final summation in the numerator instead of the R^n+1 ?

Again, thanks for this
$
\displaystyle\sum_{k=1}^n ar^k=s=ar+ar^2+...+ar^n
$

$sr=ar^2+ar^3+...+ar^{n+1}$

$s-sr=s(1-r)=ar-ar^{n+1}$

$s=\frac{a(r-r^{n+1})}{1-r}$

$s=\frac{ar(1-r^{n})}{1-r}$