Calculating growth of an exponential distribution A bacterium divides in to two bacteria at time zero. The time until bacteria divides further is an exponential random variable, with $\lambda=1$. What is the expected time $T_n$ when bacteria reach the number $n$?
I tried thinking along the relation between Poisson and exponential distribution but it got me even more confused. I am working on this approach now. 
Assuming there are $r$ bacteria at the moment, and calculating the expected duration to reach the population of $(r+1)$ bacteria denoted by $T_r$. 
$T_r=\int\limits_{0}^{\infty} rte^{-t}(1-e^{-t})^{r-1}dt$.  
Therefore the expected duration for the bacteria to reach $n$ is $\sum\limits_{r=2}^{n-1}T_r$.
How can I do this?

Consider the sequence $T_3-T_2, T_4-T_3,\ldots$. Clearly $\mathbb P(T_1=0)=\mathbb P(T_2=0)=1$. Now for each $k\geqslant 2$, we have $T_{k+1}-T_k\sim\mathrm{Expo}(k\lambda)$ as the time until a bacteria division is the minimum of $k$ $\mathrm{Expo}(\lambda)$ random variables. Therefore $\mathbb E[T_{k+1}-T_k] = \frac1{k\lambda}$. Write
$$
T_n = \sum_{k=2}^{n-1} (T_{k+1}-T_k) , then
$$
\begin{align}
\mathbb E[T_n] &= \mathbb E\left[\sum_{k=2}^{n-1} (T_{k+1}-T_k)\right]\\ &= \sum_{k=2}^{n-1}\mathbb E[T_{k+1}-T_k]\\ &= \sum_{k=2}^{n-1} \frac1{k\lambda}\\ 
&=\frac1\lambda \left(-1+\sum_{i=1}^{n-1}i\right).
\end{align}
