Optimizing Brick Stacking: Minimum Energy Requirements

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Homework Statement



N bricks, each having height h and mass m, are lying on the floor in a configuration of minimal potential energy. What is the minimum energy required to put all bricks one on top of another?

Homework Equations



U=mgy

The Attempt at a Solution



As (I assume) they are all flat on the floor with minimum potential energy, I think this problem is mainly about the most efficient way to stack the bricks. Is the energy used to stack the bricks the same as the difference in potential energy from the starting position (presumably 0 J) and the ending position (mgNh)? And how does that translate into a MINIMUM ENERGY requirement?
 
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The minimum energy just means you lift each brick directly to it's final position.
(It's not really an important part of the question.)
It's also not about the most efficient stacking - you are told that they simply go on top of each other.

Draw a stack of bricks on top of each other and think about the average potential energy,
 
Ok, I understand what you're saying. So then the potential energy for any brick N would just be mgNh, correct? And am I correct in saying that the potential energy (mgNh) is also the "minimum energy required to put all bricking one on top of another" as the question asks?
 
Ah, I didn't think of the total energy of all the bricks. So would it be mghN-mgh (or of course mgh(N-1))?
 
I'm afraid I don't understand what you're getting at
 
The first (not counting the one on the floor) brick moves through height h gaining pe of mgh
The next brick moves through 2h = mg2h
The next = mg3h
and the Nth mgNh

Whats the total?
ie what is the sum of 1+2+3+4+5...N
 
Ok, so it's the sum of mghN from N=1 to N, but I'm not sure what the sum of 1+2+3+. . . +N is. Before I was just using an integral of mghN from 1 to N, but I guess that's not how you do it.
 
So. . . I believe the energy would be mgh((n^2+n)/2)?