Does any one know how to get a compact solution to the following formula?
$$\binom{n}{1} \cdot n + \binom{n}{2} \cdot (n - 1) + ... + \binom{n}{n} \cdot 1 = \sum_{k = 1}^{n}{\binom{n}{k} \cdot (n + 1 - k)}$$
The answer may be something related to $2^n$, but I currently have no clues on how to proceed.
Thanks!
On
Start with a change of index $j=n-k$. The sum we’re looking for rewrites as
$$S=\sum_{j=0}^{n-1}(j+1){n\choose n-j}=\sum_{j=0}^{n-1}(j+1){n\choose j}$$
Then consider
$$x(1+x)^n=\sum_{j=0}^n{n\choose j}x^{j+1}$$
Derive that once to get
$$(1+x)^n+nx(1+x)^{n-1}=\sum_{j=0}^n(j+1){n\choose j}x^j$$
Evaluating at $x=1$ we have
$$2^{n-1}(2+n)=S+(n+1){n\choose n}$$ $$S=2^{n-1}(2+n)-(n+1)$$
\begin{align} \sum_{k=1}^n \binom{n}{k} (n+1-k) &=\sum_{k=1}^n \binom{n}{k} (n+1) - \sum_{k=1}^n \binom{n}{k} k\\ &=(n+1)\sum_{k=1}^n \binom{n}{k} - n \sum_{k=1}^n \binom{n-1}{k-1}\\ &=(n+1)(2^n-1) - n 2^{n-1}\\ &= (n+2) 2^{n-1}-n -1 \end{align}