$C(n,2)+2×C(n,3)+3×C(n,4)+...+(n-1)×C(n,n)$
By calculating the combinations and multiplying , then simplifying i can until here:
$(n-1)[n/2 + n(n-2)/3 + n(n-2)(n-3)/8 +...+ 1]$
I thought of transforming everything in the brackets to combinations of $n-2$ or $n$ and get something like $(2^n-2)×(n-1)$ but couldnt figure out how. Thank you in advance.
$signs and use_for subscripts.$x_1$comes out as $x_1$. – saulspatz Jan 22 '21 at 19:41