Is there a closed-form solution for this summation?

Question

Context: I want to find the amount of time it takes for the sum of exponential growth to reach a given value.

$x_t=\sum\limits_{i=0}^{t}x_0(1+r)^i$

In other words, is there a formula I can use to solve for $t$?

If there is a solution, can you please also show the derivation?

J.G. · Accepted Answer · 2020-01-16T20:01:19.273

1

An obvious Ansatz is $x_t=A(1+r)^t+B$ for constants $A,\,B$ you can find from the cases $t=0,\,t=1$. You'll get$$x_t=\frac{x_0}{r}[(1+r)^{t+1}-1],$$which can also prove by induction on $t$. This equation is obviously correct when $t=0$, and if it works with $t=k$ then$$\begin{align}x_{k+1}&=\frac{x_0}{r}[(1+r)^{k+1}-1]+x_0(1+r)^{k+1}\\&=\frac{x_0}{r}[(1+r)^{k+1}(1+r)-1]\\&=\frac{x_0}{r}[(1+r)^{k+1+1}-1].\end{align}$$(@JohnWaylandBales has also noted a telescoping sum gets this result.) Solving for $t$ gives $$(1+r)^{t+1}=1+\frac{rx_t}{x_0}\implies t=\frac{\ln\left(1+\frac{rx_t}{x_0}\right)}{\ln(1+r)}-1.$$

edited Jan 16 '20 at 20:01

answered Jan 16 '20 at 19:37

J.G.

115,835

Thanks! Can you give a bit more detail on the inductive part @J.G.? – Cameron Jan 16 '20 at 19:43
It can also be proved directly by noticing that the sum $(1+r)x_t-x_t$ is a telescoping sum. – John Wayland Bales Jan 16 '20 at 19:58
@Cameron See edit. – J.G. Jan 16 '20 at 20:01
If $\sum_{k=0}^{n-1} r^k = \frac{1-r^n}{1-r}$ then $\sum_{k=0}^{n} r^k = \sum_{k=0}^{n-1} r^k + r^n = \frac{1-r^n}{1-r}+r^n= \frac{1-r^{n+1}}{1-r}$. – marty cohen Jan 16 '20 at 20:02

Is there a closed-form solution for this summation?

1 Answers1