I need to prove $k\binom nk=n \binom{k-1}{ n-1}$ where $n$ and $k$ are integers with $1\leq k\leq n$ using an algebraic proof.
I solved the left side which is $\binom nk$ using the pascals identity for $\binom nr$. But,now when i am trying do the same for the right side, i cant solve it. Can anyone help me with this please.
For $\displaystyle n\geq k>0$ and $n, k\in \mathbb{Z}$,
$$\binom{n}{k}=\frac{n!}{(n-k)!\cdot k!}=n\cdot\frac{(n-1)!}{k\cdot (k-1)![(n-1)-(k-1)]!}=\frac nk\binom{n-1}{k-1}$$
The usual definition of binomial coefficients $\binom nk$, with only the condition $k\in\mathbf N$, is $$ \binom nk=\frac{n(n-1)\ldots(n-k+1)}{k!} =\frac{n(n-1)\ldots(n-k+1)}{k\times (k-1)\times\cdots\times 1} $$ Splitting off the initial factors of numerator and denominator for $k>0$ gives the identity $$ \binom nk=\frac nk\binom{n-1}{k-1} \qquad\text{for $k>0$.} $$ This gives after multiplying by $k$ your (presumably) desired identity $k\binom nk=n\binom{n-1}{k-1}$. This is therefore valid under the sole hypothesis that $k$ is a positive integer. If one extends the above definition by setting $\binom nk=0$ whenever $k$ is a negative integer (again this is conventional), then it even holds for all $k\in\mathbf Z$.