Its this : (Need to solve the Arithmetico-Geometric Progression and its worser counterpart ) $$\frac{n}{2}\sum_{k=0}^n k\cdot2^{k} - \frac{1}{2}\sum_{k=0}^n k^{2}\cdot2^{k}$$Which boils down to : $$\frac{n}{2}[(n-1)\cdot2^{n+1} + 2] - \frac{1}{2}[2^{n+2} + (n-1)^2\cdot2^{n+1}-6]$$Which condenses to (if I have'nt got the arithmetic wrong) $$(n-3)\cdot2^{n}+n+3$$

Lets take a simplified case of $n = 10$ and generalise while solving it : ( We are neglecting the singleton sets and $\phi$ because they dont yield anything to the answer ) Say $(x , y) = (2 , 8)$ we want a $f(x,y)$ which gives the number such subsets (where x and y are the minimum and the maximum elements) of the orginal set ($U$) The number of such subsets is just the number of subsets of : $(3,4,5,6,7)$ ( This contains $(y-x-1)$ elements ) (Since all other elements can not exceed 8 or preceed 2) Hence We can create a set $X$ with $(2 , 8)$ and all the other the other elements of the subset of $(3,4,5,6,7)$ The number of such subsets are thus : $f(x,y) = 2^{y-x-1}$ and contributes $(y-x)\cdot2^{y-x-1}$ to the sum It can be trivially solved that $ y-x = k$ has $(n-k)$ solutions when $x\hspace{1mm}, y$ vary from $(1,2,3,...,n)$ Hence our final solution is : ( Since k = 0 and k = n this diminishes I wrote it orginally as such) $$\sum_{k\hspace{1mm}=\hspace{1mm}1}^{n-1} (n-k)\cdot k\cdot2^{k-1}$$