How can I solve C(n,2)+2×C(n,3)+3×C(n,4)+...+(n-1)×C(n,n)=? 'm able to find an answer but I don't think it's the right one. I tested it for n=2 and it worked but not for n=3.Here's my answer:
$$C(n,2)+2×C(n,3)+3×C(n,4)+...+(n-1)×C(n,n)=[C(n,2)+C(n,3)+...+C(n,n)] + [0×C(n,2)+1×C(n,3)+2×C(n,4)+...+
(n-2)×C(n,n)] =
2^{(n-2)} + 2^{(n-2)}×(n-2)$$
I'm either forgetting something or it's just fully wrong. Sorry for the bad formatting and thank you in advance
 A: $$\sum_{r=2}^n(r-1)\binom nr=\sum_{r=2}^nr\binom nr-\underbrace{\sum_{r=2}^n\binom nr}_{(1)}$$
Now for $r\ge1,$ $$r\binom nr=r\cdot\dfrac{n\cdot(n-1)!}{r\cdot(r-1)!\cdot(n-1-(r-1)))!}=n\binom{n-1}{r-1}$$
$$\implies\sum_{r=2}^nr\binom nr=n\sum_{r=2}^n\binom{n-1}{r-1}$$
$$\sum_{r=2}^n\binom{n-1}{r-1}=-\binom{n-1}0+\underbrace{\sum_{m=0}^{n-1}\binom{n-1}m}_{(2)}$$
From $(1),$ $$\sum_{r=2}^n\binom nr=-\binom n0-\binom n1+\underbrace{\sum_{r=0}^n\binom nr}_{(3)}$$
Use $\displaystyle(1+1)^m=\sum_{k=0}^m\binom mk$ for $(2),(3)$
A: Note that
\begin{align*}
n(1 + x)^{n - 1}  - (1 + x)^n & = \left( {\frac{d}{{dx}}(1 + x)^n } \right) - (1 + x)^n  = \sum\limits_{k = 0}^n {\binom{n}{k}kx^{k-1} }  - \sum\limits_{k = 0}^n {\binom{n}{k}x^k } \\ & = \sum\limits_{k = 0}^n {\binom{n}{k}(k - x)x^{k-1} } .
\end{align*}
Now just substitute $x=1$.
A: You could take a combinatorial approach. As usual, let $[n]=\{1,2,\ldots,n\}$.
$(k-1)\binom{n}k$ is the number of ways to choose a $k$-element subset of $[n]$ and then mark any one of its elements except the largest. Summing from $k=2$ to $k=n$ gives you the number of ways to choose any subset of $[n]$ with at least $2$ elements and then mark any one of its elements except the largest; that’s the sum that you’re trying to evaluate.
Alternatively, you could carry out the same task by the following procedure.

*

*Mark any $k\in[n]$; you can do this in $n$ ways.

*Pick any subset $A$ of the remaining $[n]\setminus\{k\}$ to get a set $S=A\cup\{k\}$ with one marked element; you can do this in $2^{n-1}$ ways, so there are $n2^{n-1}$ ways to form a subset $S$ of $[n]$ with one marked element.

*If $k=\max S$, discard $S$, since its marked element is its largest element. By construction $S$ could be any non-empty subset of $[n]$; there are $2^n-1$ such subsets, so we’re discarding $2^n-1$ improperly marked sets $S$.

In the end we have $n2^{n-1}-\left(2^n-1\right)=(n-2)2^{n-1}+1$ suitably marked subsets of $[n]$, so
$$\sum_{k=2}^n(k-1)\binom{n}k=(n-2)2^{n-1}+1\,.$$
For $n=4$, for instance, we have
$$\binom42+2\binom43+3\binom44=6+8+3=17=2\cdot2^3+1\,.$$
