Geometric Progression Problem

[Guest]

Hello everyone,

I attempted to find the solution for Problem Number 2. My solution was incorrect.

I reviewed the solution that they offered, and I understand every step except where it states in the 5th row

an+1−a1=3.(1+3+⋯+3n−1)=3.3n−13−1=3n+1−32.

My specific question is what happens after we simply the series at "(1+3+⋯+3n−1)" into "3.3n−13−1"(Please view the link below to get an accurate description of the problem)? It looks similar to the sum of n terms formula "Sn=a(1-r^n)/1-r". When I do plug in the formula my numbers are switched, and I get a negative value for when I solve for a10. Any input would be very appreciated!

http://www.math10.com/problems/geometri ... difficult/

-Jason

[Guest]

[tex]1+3+3^2+...+3^{n-1}[/tex] is a geometric series (which you correctly identified).
The formula for the geometric series [tex]1+r+r^2+...+r^{n-1}[/tex] is
[tex]\frac{r^{n}-1}{r-1}[/tex] or [tex]\frac{1-r^{n}}{1-r}[/tex]
they are the same formula the only difference is a factor of -1 in both the numerator and denominator.
(It should be obvious that -p/-q = p/q.)
Using this formula [tex]1+3+3^2+...+3^{n-1} = \frac{3^{n}-1}{3-1}[/tex] or if you prefer [tex]\frac{1-3^{n}}{1-3}[/tex].


The actual sum they are interested in is [tex]3+3^2+3^3...+3^{n}[/tex] they compute this sum by first factoring out 3 to get [tex]3\times(1+3+...+3^{n-1})[/tex] and then replacing the bracket with our geometric formula:
[tex]3\times\left(\frac{3^{n}-1}{3-1}\right)[/tex] or [tex]3\times\left(\frac{1-3^{n}}{1-3}\right)[/tex]
You could also get to the same result by using the more general form of the geometric series formula:
[tex]a+ar+ar^2+ar^3+...+ar^{N-1} = a\left(\frac{r^{N}-1}{r-1}\right)[/tex] or equivalently [tex]a\left(\frac{1-r^{N}}{1-r}\right)[/tex]
In our case [tex]a=3[/tex], [tex]r=3[/tex], and [tex]N=n[/tex] substituting this into the formula gives us the same expressions as before.


To calculate [tex]a_{10}[/tex] we first find a formula for [tex]a_n[/tex] and to do that we first find a formula for [tex]a_{n+1}[/tex].
[tex]a_{n+1} -a_1 = 3+3^2+3^3...+3^{n} = 3\times\left(\frac{1-3^{n}}{1-3}\right)[/tex]
the right hand side simplifies
[tex]3\times\left(\frac{1-3^{n}}{1-3}\right) = 3\times\left(\frac{1-3^{n}}{-2}\right)[/tex]
[tex]=\frac{3\times\left(1-3^{n}\right)}{-2}[/tex]
[tex]=\frac{3-3^{n+1}}{-2}[/tex]
We know that [tex]a_1=1[/tex] so our formula becomes
[tex]a_{n+1}-1=\frac{3-3^{n+1}}{-2}[/tex]
which rearranges to
[tex]a_{n+1}=\frac{3-3^{n+1}}{-2}+1[/tex]
Substituting [tex]n = N-1[/tex] into the formula tells us that
[tex]a_{(N-1)+1} = \frac{3-3^{(N-1)+1}}{-2}+1[/tex]
which simplifies to
[tex]a_{N} = \frac{3-3^N}{-2}+1[/tex]
Substituting [tex]N=10[/tex] tells us that
[tex]a_{10}=\frac{3-3^{10}}{-2}+1[/tex]
[tex]=\frac{3-59049}{-2}+1 = \frac{-59046}{-2}+1 = 29523+1 = 29524[/tex]
The same answer as given in the link.


Hope this helped,

R. Baber.