A brief question: I know that $\binom{n}{k}$ isn't defined for $n<k$, however, can I show that $n\cdot \binom{n-1}{n}=0$? if so how can I do that? I saw through wolfram that it really is zero, and I was wondering how he got into it. I have encountered this debate in a question when I needed to find a closed form of the next series: $$\sum _{k=0}^{n}\binom{n}{k}( n-k)$$
And instead of getting $n$ out first, I have tried to do the following steps: $$\sum _{k=0}^{n}\binom{n}{k}( n-k) =\sum _{k=0}^{n}\frac{n!}{k!( n-k) !}( n-k) =\sum _{k=0}^{n}\frac{n!}{k!( n-k-1) !}\\ \\ =n\sum _{k=0}^{n}\frac{( n-1) !}{k!( n-k-1) !} =n\sum _{k=0}^{n}\binom{n-1}{k} =n\left[\binom{n-1}{n} +\underbrace{\sum _{k=0}^{n-1}\binom{n-1}{k}}_{ \begin{array}{l} \text{by Newton's binomial}\ \\ \text{it is }2^{n-1} \end{array}}\right]\\ =n\left[\binom{n-1}{n} +2^{n-1}\right] =n\binom{n-1}{n} +n2^{n-1}$$
Now is it valid to right this as final result of a closed form for $\sum _{k=0}^{n}\binom{n}{k}( n-k)$?
I will be happy to see from you if both questions I wrote are correct, and why.
$n\choose k$ is equal to $0$ for $n<k$, so your expression simplifies to $$n\cdot 0$$ which is obviously equal to $0$.
The fact that $${n\choose k} = 0$$ for $n<k$ comes down to the definition of $n\choose k$. The value is defined as
From this definition, the fact should be fairly obvious, since if $k>n$, there are no subsets of an $n$-element set that contain more than $n$ elements.