Let
$(a+b+z)^n = A$
Now, I'm interested in the term $abz$ in $A$. ($a,b,z$ can have different and varying exponents)
So I try to expand the left side using the Binomial Theorem.
$(x+y)^n = c_0 x^n y^0+c_1x^{n-1}y^1+c_2x^{n-2}y^2+\ldots+c_{n-1}x^1y^{n-1}+c_nx^0y^n$
where $x =a$ and $y = b + z$
$(a+(b+z))^n = c_0 a^n (b+z)^0+c_1a^{n-1}(b+z)^1+c_2a^{n-2}(b+z)^2+\ldots+c_{n-1}a^1(b+z)^{n-1}+c_na^0(b+z)^n$
If $n=2$, the $abz$ term vanishes.
$(a+(b+z))^2 = c_0 a^2 (b+z)^0+c_1a(b+z)^1+c_2a^{0}(b+z)^2$
$(a+(b+z))^2 =c_0 a^2 +c_1a(b+z)+c_2(b+z)^2$
If $n=3$, the $abz$ term is present but the requirement is that $n$ must be very close to 2. ($n$ approaches 2 from the right)
$(a+(b+z))^3 = c_0 a^3 (b+z)^0+c_1a^{2}(b+z)^1+c_2a^{1}(b+z)^2+c_3a^0(b+z)^3$
$(a+(b+z))^3 = c_0 a^3 +c_1a^{2}(b+z)+c_2a(b+z)^2+c_3(b+z)^3$
$(a+(b+z))^3 = c_0 a^3 +c_1a^{2}(b+z)+c_2a(b^2+2bz+z^2)+c_3(b+z)^3$
$(a+(b+z))^3 = c_0 a^3 +c_1a^{2}(b+z)+c_2a(b^2+z^2)+ 2c_2abz+c_3(b+z)^3$
So for $n=3$:
(term with abz)/A =$ \frac{2c_2abz}{c_0 a^3 +c_1a^{2}(b+z)+c_2a(b^2+z^2)+ 2c_2abz+c_3(b+z)^3}$
So for $2+\delta>n>2$, ($\delta$ is very small):
(term with abz)/A = $\frac{c_{n-1}a^1(b+z)^{n-1}}{c_0 a^n +c_1a^{n-1}(b+z)+....+c_{n-1}a^1(b+z)^{n-1}+c_n(b+z)^n}$
Or:
(term with abz)/A = $\frac{c_2a^{n-2}(b+z)^2}{c_0 a^n+c_1a^{n-1}(b+z)+....+c_{n-1}a^1(b+z)^{n-1}+c_n(b+z)^n}$
Question: How large is the term abz in $(a+b+z)^n$ as $n$ approaches 2?
Calculating the limit:
$\lim\limits_{n \to 2} \frac{c_{n-1}a^1(b+z)^{n-1}}{c_0 a^n +c_1a^{n-1}(b+z)+....+c_{n-1}a^1(b+z)^{n-1}+c_n(b+z)^n}$ $=\frac{c_{1}a^1(b+z)^{1}}{c_0 a^2 +c_1a^{1}(b+z)+c_2(b+z)^2}$