So this what the combination series looks like,
$\sum_{n=0}^{n_B-1} \binom{n_B-1}{n}\left((a)^{n_B-1-n} + (b)^{n}\right)$
The value of $n_B \in \{2,3,\dots N\}$ and is fixed. For different values of $n_B$, the series looks as,
$n_B = 2$, $\sum_{n=0}^{1} \binom{1}{n}\left((a)^{1-n} + (b)^{n}\right) = a+b$
for $n_B = 3$, $\sum_{n=0}^{2} \binom{2}{n}\left((a)^{2-n} + (b)^{n}\right) = a^2+2(a+b)+b^2$
for $n_B = 4$, $\sum_{n=0}^{3} \binom{3}{n}\left((a)^{3-n} + (b)^{n}\right) = a^3+3(a^2+b)+3(a+b^2)+b^3$
and so on, somewhat similar to $(a+b)^{n_b -1}$, but with terms like $2(ab)$ become $2(a+b)$ in $n_B = 2$ and so forth in other values of $n_B$.
Now, is there any way I can deduce this series to remove the summation?. Any inputs would be helpful
To remove unnecessary complication, set $k=n_B-1$. The series in question is then $$ \sum_{n=0}^k \binom kn \bigl( a^{k-n} + b^n \bigr) = \sum_{n=0}^k \binom kn 1^na^{k-n} + \sum_{n=0}^k \binom kn b^n1^{k-n} = (a+1)^k + (b+1)^k. $$ Note that the evaluations in the OP are all missing the constant term $2$, but are otherwise special cases of this formula.