Question: What is the advantage for rockets to [have multiple stages](http://en.wikipedia.org/wiki/Multistage_rocket)? 

Wouldn't a single stage with the same amount of fuel weigh less?

**Note** I would like a quantitative answer, if possible :-)

![alt text](https://i.stack.imgur.com/lswdV.jpg)

Answer: 
**Edited a little now that I better understand your question.**

**Short**  

In a multi-stage, the weight of the parts that are dropped along the ride compensates for the fact that the extra engines make it heavier in the beginning. Partially because a rocket's engine isn't that heavy compared to the fuel tank. The engine mostly just ignites and controls the combustion, while the fuel tank needs to be huge.

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**Long**  

I'll analyse the very optimal case, in which the energy spent is the minimum energy needed to get your rocket in orbit. Real cases go waaay above this and it's not even practical to do energy balance. [Thornton & Marion](http://rads.stackoverflow.com/amzn/click/0534408966) does a great (undergraduate level) analysis using linear momentum.

In order to launch a single stage rocket, there's a specific size that minimizes the needed energy. The weight of your rocket (just the hull, not the fuel) increases with size $P\_R=g\rho V$. And the minimum amount of energy necessary to get your rocket in orbit obviously increases with its weight: $$W\_{min} = hP\_R + W',$$ where $W'$ is the work you do to lift the fuel. $W'$ is not important here but, for the record, it goes something like:
$$W' = g\int\_0^h m\_{fuel}(z)dz'.$$

So, if the rocket is too small, the little amount of fuel you can fit into it does not contain enough energy to lift the hull. If the rocket is too big, then you're just wasting energy because it's unnecessarily heavy. That means you have an optimal size which minimizes the energy of a single-stage rocket. Once you find it, the amount of fuel associated to that size is the very minimum you'll need to get your rocket on orbit.

Now consider a rocket of that same size, but with 2 stages. Let's assume that it has an extra engine of weight $p\_e$. Let's also assume that, when your rocket reaches a height $h/2$ more than $1/3$ of its tank will be empty (which is true). Since that part of my tank is empty, I might as well leave it behind (along with the engine) and drop the dead weight (I'll call it $p\_t$). If I drop it at $h/2$ the energy I'll need to reach orbit is
 $$W\_{min2} = \frac{h}{2}(P\_R + p\_e) + \frac{h}{2}(P\_R - p\_t) + W'\_2 = hP\_R + W'\_2 - \frac{h}{2}(p\_t - p\_e).$$
If we sightly reduce the amount of initial fuel we can make $W'\_2\leq W'$. Therefore, $W\_{min2}$ is clearly less than $W\_{min}$ as long as the engine you had to add is lighter than the tank you dropped.
Launch weight would be lower if you had a fixed fuel load, and only one engine and fuel tank. However the specific impulse applied to the payload would be lower. The problem is that even when the fuel is say 90% exhausted, the rocket is still trying to accelerate the now grossly oversized fuel tank and engine. So the trick is to try to reduce the deadweight (structural mass) as the fuel is consumed. Another compromise system is to have jetisonable external fuel tanks, like the shuttle, which are thrown away once their fuel is consumed.