Sum of all the not empty subsets

[Marina19]

Hello!

I've got a problem for which I have worked out the asnwer however I have some problems with writting the solution for my classes.

There is a given integral \[n\geq 1\]. For not empty subset of the set X {1,2,...,n} let x and y be respectively the smallest and the biggest element of the set X and let
\[f(X)=\frac{1}{n-(y-x)}\]
Find the sum of the f(X) for all not empty subsets X of the set {1,2,...,n}

Solution:

First consider 1-element subsets:
so we have n of them and the difference (y-x)=0 so we have n*(1/n)=1
Now consider subsets for which the difference y-x=1
so we have n-1 of them and then f(X)=(n-1)/(1/(n-1))
Now for y-x=2 we have 2*(n-2)*(1/(n-2))

And for the y-x=k we've got
\[f(X)=(n-k)\frac{1}{n-k}2^{k-1}\]

So now let's call the sum of all the subsets S(n)
\[S(n)=2^{n-1}
\]
as S(1)=1, S(2)=2, S(3)=4 and each S(n)=S(n-1)+2^(n-2)

[Corei13]

সমাধানটা ঠিকই তো আছে।

[Marina19]

Sorry but I don't understand it and neither does google translate However, what I would like is some adivce how to write this solution so it will be clear to understand

[Corei13]

Oops! Sorry, I thought you're from Bangladesh!
OK, here's your solution:
Let, $y-x=d > 0$ then, there are $2^{d-1}$ subsets of $\{1,2,...,n\}$ with maximum and minimum element $y$ and $x$, respectively, because, there're $d-1$ elements between $x$ and $y$, and for each such element, we have two choice, either take it or leave it.
Now, $1 \leq x=y-d \leq n-d$, so, we have $n-d$ values for $x$, and for each such $x$ we have $2^{d-1}$ subsets.
Now, take $d=1$ to $n-1$ and add all $F(X)$'s. The summation is $\sum_{i=1}^{n-1}{\left ( (n-d) \cdot 2^{d-1} \cdot \frac{1}{n-d} \right )}=2^{n-1} -1$
Now, let, $x=y$. So, they are one element subset and there are $n$ such subsets. So, we have to add another $n\cdot \frac{1}{n-0} = 1$
So, the answer is $2^{N-1}$