# Recursive to Sum if possible

• Mar 26th 2011, 02:35 PM
dwsmith
Recursive to Sum if possible
Given the following recursive definition is it possible to write it as a sum?

$\displaystyle a_0=1,337,101.43 \ \ \ a_n=a_{n-1}(1+33*.02)*1.03^{n-1}$
• Mar 26th 2011, 03:59 PM
HallsofIvy
So $x_n= (1.66)(1.03)^{n-1}a_{n-1}$?

My suggestion would be to calculate some of them:
$a_1= (1.6)a_0$
$a_2= (1.6)(1.03)a_1= (1.6)^2(1.03)a_0$
$a_3= (1.6)(1.03)^2a_2= (1.6)^3(1.03)^3a_0$
$a_4= (1.6)(1.03)^3a_3= (1.6)^4(1.03)^6a_0$
$a_5= (1.6)(1.03)^4a_4= (1.6)^5(1.03)10a_0$
etc.

So it's clear that the formula involves $(1.6)^{n-1}$. The powers on 1.03 are 0, 1, 1+ 2= 3, 1+ 2+ 3= 6, 1+ 2+ 3+ 4= 10, etc. for n= 1, 2, 3, 4, 5, ... That is, of course, $\frac{n(n- 1)}{2}$.