How $\sum\limits_{k=0}^n {n\choose k}\cdot 0^{n-k}\cdot 2^{k}=2^{n}$?

Question

Note: I know this can be converted to $(0+2)^{n}=2^{n}$ according to the binomial theorem, I just got stuck after expanding the sum.

I expanded this sum and this is what I got:

$0+0+0+...+{n\choose n}\cdot 0^{0}\cdot 2^{n}$

Here, since $0^{0}$ is undefined how the sum will be equal to $2^{n}$ ?

Answer 1

Use $0^0=1, 0^k=0, k\ne 0$ In the series on ly the last term contibutes : $$(0+2)^n=0+...+0+...+0+...+0+{n \choose n} 0^{n-n} 2^n=2^n.$$

Answer 2

Indeed, it's necessary to define $0^0 = 1$ here.

Note that there are cases where it's not right to define $0^0 = 1$; for example, $\lim_{x \to 0} 0^x$ is $0$. For this reason, anywhere you have a formula which includes a term $0^0$, you need to give the value of $0^0$ alongside that formula.

Answer 3

It's necessary to define $0^0 = 1$ only if you use $(a+b)^n=\sum_{k=0}^n {n\choose k}a^{n-k}b^{k}$.

The issue does not arise if you use $(a+b)^n=a^n+{n\choose 1}a^{n-1}b+\cdots +{n\choose n-1}ab^{n-1}+b^n$.