Simplify mixed arithmetic/geometric progression

[edcrane]

I need to solve something that looks like a mix between an arithmetic and geometric progression. Solving a plain old arithmetic progression should be easy like:

S = a + 2a + 3a +...+na
=> S(2) = (na + a) + (n-1(a)+2a) + (n-2(a)+3a)+...(1(a)+na)
=> S = n(n+1)a/2

And likewise, I get the standard S(1-r) technique for simplifying geometric progressions, but now I'm given this:

S = r + ar^2 + 2ar^3 +...+(n-1)ar^n

What's the strategy for simplifying this one?

 

[pka]

First, have you left off an 'a' in the first term?
It looks like a derivative of a finiite geometric series that has been multiplied by ar.

 

[edcrane]

First, have you left off an 'a' in the first term?
It looks like a derivative of a finiite geometric series that has been multiplied by ar.

no, there's no a in the first term. I thought the first step might be to divide S by r, s.t.:

S(1/r)=1 + ar + 2ar^2 +...(n-1)ar^(n-1)

Do I need to be integrating both sides?

 

[edcrane]

Ok, here's what I have:

S(1/r)-1= ar + 2ar^2 +...(n-1)ar^(n-1)

=>[S(1/r)-1](1/ar)=1+2r+3r^2+...+(n-1)r^(n-2)

**Integrate both sides**

=>(-S/ar -ln(r)/a)=r + r^2 +r^3 +...+r^(n-1)

=>(-S/ar -ln(r)/a)(1-r)=r-r^n

=>S=[(r^n-r)/(1-r)-ln(r)/a](ar)

Does this look right?

 

[pka]

Here is a start:
\(\displaystyle \L\begin{array}{l}
T = a + ar + ar^2 + \cdots + ar^{n - 1} = \frac{{a - ar^n }}{{1 - r}} \\
T' = a + 2ar + 3ar^2 + \cdots + a(n - 1)r^{n - 2} = \frac{{a - ar^n - nar^{n - 1} + anr^n }}{{\left( {1 - r} \right)^2 }} \\
r^2 T' = ar^2 + 2ar^3 + \cdots + a(n - 1)r^n = \frac{{ar^2 - ar^{n + 2} - nar^{n + 1} + anr^{n + 2} }}{{\left( {1 - r} \right)^2 }} \\
\end{array}\)