Suppose that $x_1<x_2<\ldots<x_n$ are our real numbers. Then $x_1+\ldots+x_k<x_1+\ldots+x_{k-1}+x_{k+1}<\ldots<x_1+\ldots+x_{k-1}+x_n$ (we repeatedly increased the last indice) After that $x_1+\ldots+x_{k-1}+x_n<x_1+\ldots+x_{k-2}+x_{k}+x_n<\ldots<x_1+\ldots+x_{k-2}+x_{n-1}+x_n$ (we repeatedly increased the -one to the last- indice) As above we now repeatedly increase the -two to the last- indice and so on. The last step is $x_1+x_{n-k+2}+\ldots+x_n<x_2+x_{n-k+2}+\ldots+x_n<\ldots<x_{n-k+1}+x_{n-k+2}+\ldots+x_n$. One can see that at each step we are introducing $n-k$ new numbers except the first step for which we also introduce $x_1+x_2+\ldots+x_k$. So the numbers mentioned above are $k(n-k)+1$ numbers as desired.

Assume $x_{1}<x_{2}<...<x_{n}$(*) All set T have $k(n-k)+1$:i in[1,n-k], j in[1,k]. $b_{ij}=(\sum_{u=0}^{k} (x_{i+u}))-x_{i+1+k-j}$ Easy to see that: $b_{ij}<b_{i(j+1)}$ $b_{ik}<b_{(i+1)1}$ because that hence all $b_{ij}$ are dictint. Have k(n-k) number $b_{ij}$ and have a $a_{max}=x_{n}+x_{n-1}+...+x_{n-k+1}$ greater than all $b_{ij}$ All for us QED

Here's a visual solution. Let $x_1 < x_2 < ... < x_n$ be the elements of $S$, and call a subset of $S$ of size $k$ a k-set for short. Place marbles at locations $(1, 2, ..., k)$ on the number line. In a move, we are allowed to select a marble at position $m$ and slide it one step to $m+1$, so long as $m+1$ is not currently occupied by a marble. It is not hard to see that: (i) We can make a sequence of moves that results in the marbles being at $(n-k+1, ..., n)$. Call this the final state. (ii) Each move increases the sum of occupied positions by exactly one. Hence, exactly $k(n-k)$ moves are made to move from initial to final state. (iii) Let each configuration of marbles correspond to a k-set in the natural way. Then each move strictly increases the sum of the elements of the currently represented k-set. Hence our process generates $k(n-k) + 1$ distinct k-sets with distinct sums. $\blacksquare$

This is easy by induction n

$\color{magenta} \boxed{\textbf{SOLUTION C2}}$ Let, $$x_1 < x_2 < x_3 < ... < x_n \in S$$ We need to show there are atleast $k(n-k)+1$ distinct elements of the form $x_1 + x_2 + ... + x_k$ In each step,We turn $x_i \implies x_{i+1}$ and after $(n-k)$ steps we will get the maximum sum $x_{n-k+1} + x_{n-k+2} + ... +x_n$ In each sub-step, we turn one $x_j \in (x_1,x_2,...x_k) \implies x_{j+1}$ untill maximum sum is achieved. We see in each step we have $k$ sub-steps for each $(x_1,x_2,...x_k)$ And all the sums are strictly increasing as $x_i < x_{i+1}$ for all $i \in (1,2,3...n-1)$ So, We get $k(n-k)$ distinct sums in $(n-k)$ steps and we had $x_1+x_2+...x_k$ atfirst. Therefore, $T$ contains atleast $k(n-k)+1$ distinct elements $\blacksquare$