Display cans of food in a square-based pyramid. Full question:
The manger of another grocery store asks a stock clerk to arrange a display of canned vegetables in a square-based pyramid (top is one can, 4 cans under then, 9 cans under top 2 levels, etc) How many levels is the tallest complete pyramid that we can make with 410 cans of vegetables? How many cans are left over as extras after the complete pyramid has been made?
Im trying to figure it out but not sure of the pattern or if there is a formula to calculate it?
Any tips or solutions would help because I am lost!
Thank you,
-Trap
 A: nostafict gave you the way to go. Now, if you know that the sum of the squares of the first $n$ natural numbers is (Faulhaber's formula) $$S=\sum_{i=1}^n i^2=\frac{1}{6} n (n+1) (2 n+1)$$ you just need to find $n$ such that $S$ be the closest to $410$.
From an algebraic point of view, you could try to solve the corresponding cubic equation and obtain that the only real solution of $$\frac{1}{6} x (x+1) (2 x+1)=410$$ is $x=10.2222$, so $n=10$.
You could also try to find an upper bound of the solution noticing that $S > \frac{1}{6} n (n) (2 n)=\frac{n^3}{3}$ which then implies that $n^3<3\times 410=1230$ that is to say $n<10.7144$ and by inspection find that $n=10$ is the solution.
Now, you could play with numbers larger than $410$ using the same method. Graphing the function could also help.
A: Number of cans goes like.. 1, 4, 9....n, the total of which should be 410
I.e. 1 squared + 2 squared + 3 squared... + n squared = 410
Sum of number squared series
Solve it.. 
P.S. Bear with me not being able to write notations, using my phone right now
