Closed form expressions for $\sum_{k=0}^n {n\choose k}b^{k}$ and $\sum_{k=0}^n {n\choose k}b^{n-k}$?

Question

I know that $\sum_{k=0}^n \binom{n}{k}~b^{k}=(b+1)^n$, because the terms in the binomial expansion of $(b+1)^n$ are exactly the terms in the summation.

Could we conclude that $\sum_{k=0}^n \binom{n}{k}~b^{n-k}=(b+1)^n$ as well, if we instead make an expansion on $(1+b)^n$? Does this make any sense as an explanation?

Answer 1

$$LHS=\sum_{k=0}^n {n\choose k}b^{n-k}=\sum_{k=0}^n {n\choose n-k}b^{n-k}$$

Now let $k'=n-k$

$$LHS=\sum_{k'=0}^n {n\choose k'}b^{k'}=(1+b)^n$$

Answer 2

Just distribute the factor of $b^n$ out from the series, apply the binomial expansion, then distribute it back inside.

$$\begin{align}\sum_{k=0}^n\binom nk b^{n-k} &= b^n\sum_{k=0}^n \binom nk (b^{-1})^k\\[1ex]&=b^n(b^{-1}+1)^n\\[1ex]&=(1+b)^n\end{align}$$