# Is it possible to find the nth term in this sequence?

$A_{n=0}=A_0e^{-\alpha d}$

$A_1=A_0+\rho A_0e^{-2\alpha d}$

$A_2=A_0+\rho(A_0+\rho A_0e^{-4\alpha d})$

$A_3=A_0+\rho[A_0+\rho(A_0+\rho A_0e^{-8\alpha d})]$

$A_4=A_0+\rho[A_0+\rho[A_0+\rho(A_0+\rho A_0e^{-16\alpha d})]]$

You probably get the idea of what is going on. It's basically $A_n=A_0+\rho A_{n-1}$, except in the $A_{n-1}$ we have a different exponential, namely the exponent from the previous one multiplied by 2.

I decided to ignore $A_{n=0}$ and try to find the nth value of the sequence starting from $n=1$.

I decided to try and make it easier by expanding and factorising the brackets in a neater way:

$A_1=A_0(1+\rho e^{-2\alpha d})$

$A_2=A_0(1+\rho +\rho^2e^{-4\alpha d})$

$A_3=A_0(1+\rho+\rho^2+\rho^3e^{-8\alpha d})$

$A_3=A_0(1+\rho+\rho^2+\rho^3+\rho^4e^{-16\alpha d})$.

So this allows me to obtain

$A_n=A_0(1+\rho^ne^{-2n\alpha d}+\rho^{n-1}+\rho^{n-2}+\rho^{n-3}+...)$,

and this carries on until the number of lone rho values are equal to one less than the value of n. I could not for the life of me find a sequence that satisfies these lone rhos.

Any help?

• If your manipulations are correct, I don't see what more you want. Your expression for $A_n$ is a perfectly good one. – Patrick Stevens Oct 7 '15 at 7:14
• I don't want a "..." at the end. – ODP Oct 7 '15 at 8:31
• $$A_n/A_0 = \rho^n e^{-2n \alpha d} + \sum_{i=0}^{n-1} \rho^i$$ – Patrick Stevens Oct 7 '15 at 11:21