Answer:
How many cans will be on the bottom layer, and will there be any cans left over?
On the bottom layer will be 31 cans  and there will be 4 cans left over
Step-by-step explanation:
To resolve this problem we need to know Gauss sum of natural numbers, from k = 1 to n:
∑[tex]k=\frac{n*(n+1)}{2}[/tex] (1)
We want to know the value of n when the sum is equal or close to 500, so we replace 500 in the equation (1) to find n:
[tex]500=\frac{n*(n+1)}{2}[/tex]
[tex]500*2=n*(n+1)[/tex]
[tex]1000=n^2+n[/tex]
[tex]n^2+n-1000=0[/tex] (2)
We need to use the quadratic formula to resolve the equation (2)
[tex]n=\frac{-b+/-\sqrt{b^{2} -4ac} }{2a}[/tex] (3)
The coefficients according to equation (2) are:
a=1
b=1
c= 1000
Now, using the quadratic formula we can find n :
Positive sign before the square root,
[tex]n_1=\frac{-1+\sqrt{1^{2} -4(1)(-1000)} }{2(1)}[/tex]
[tex]n_1=\frac{-1+\sqrt{1 + 4000} }{2}[/tex]
[tex]n_1=\frac{-1+63.25}{2}[/tex]
[tex]n_1=\frac{62.25}{2}[/tex]
[tex]n_1=31.13[/tex]
Negative sign before the square root,
[tex]n_2=\frac{-1-\sqrt{1^{2} -4(1)(-1000)} }{2(1)}[/tex]
[tex]n_2=\frac{-1-\sqrt{1 + 4000} }{2}[/tex]
[tex]n_2=\frac{-1-63.25}{2}[/tex]
[tex]n_2=\frac{-64.25}{2}[/tex]
[tex]n_2=-32.13[/tex]
The negative term doesn't have sense in this case, We will use the positive term, the whole number closest to the answer we got when clearing n is 31, we will replace n=31 in the Gauss sum of natural numbers equation. (1)
∑[tex]=\frac{n*(n+1)}{2}[/tex]
∑[tex]=\frac{31*(31+1)}{2}[/tex]
∑[tex]=\frac{31*(32)}{2}[/tex]
∑[tex]=\frac{992}{2}[/tex]
∑[tex]=496[/tex]
Now, we know the triangular pyramid will be 31 cans on the bottom layer and there will be 4 cans left over (500-496=4)
