Setting vertices to be $f_0$, edges $f_1$, facets $f_2$, a tetrahedron has the $f$-vector $f=(f_0,f_1,f_2)=(4,6,4)$. When solving the Dehn-Sommerville equations for $k=0,...,d-1$
$$\sum_{j=k}^{d-1} (-1)^j \binom{j+1}{k+1}f_j=(-1)^{d-1}f_k$$
for a tetrahedron, $d=3$ and this gives the following equations
$$f_0=f_0-2f_1+6f_2$$ $$f_1=-f_1+3f_2$$ $$f_2=f_2$$
the first 2 equations simplify to $f_1=3f_2$ and $2f_1=3f_2$. I'm familiar with the second equation as it gives the right result, but what to make of the 1st one? Is there an error somewhere or why does the first result come up?
There is a similar formula for a $d$-simplex, which a polyhedron also satisfies but it gives the same results for $k=0,...,d$
$$f_{k-1}=\sum_{i=k}^{d}(-1)^{d-i}\binom{i}{k}f_{i-1}$$
Well, $${{j+1}\choose{0+1}}={{j+1}\choose{1}}=\frac{(j+1)!}{1!\ \cdot\ ((j+1)-1)!}=\frac{(j+1)!}{j!}=j+1,$$ thus you got your first equation (the one for $k=0$) wrong. It ought rather be $$+f_0-2f_1+\color{red}{3}f_2=+f_0$$ and thus it reduces here to nothing but the same result as the second equation.
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Btw. you also could define $f_{-1}:=1$ and then evaluate the equation for $k=-1$. Using $${{j+1}\choose{-1+1}}={{j+1}\choose{0}}=\frac{(j+1)!}{0!\ \cdot\ ((j+1)-0)!}=\frac{(j+1)!}{(j+1)!}=1,$$ that one then gives $$-f_{-1}+f_0-f_1+f_2=+f_{-1},$$ which surely would be nothing but the well-known Euler equation $V-E+F=2$.
--- rk