TITLE: How many different values can that sum take?
QUESTION [8 upvotes]: Let $x_1,x_2,\dots,x_{100} $ be a permutation of $1,2,\dots,100.$ How many different values does the sum $ x_1+2x_2+\cdots+100x_{100}$ take?

REPLY [5 votes]: Let $x_1,x_2,\dots,x_n$ be a permutations of $1,2,\dots,n.$ $(n\ge 4)$. 
Then there are 
$$
\dfrac{n(n-1)(n+1)}{6}+1
$$
different values of sums $ x_1+2x_2+\cdots+nx_n$.
Smallest value is 
$$v_n = 1\cdot n + 2(n-1)+3(n-2)+\cdots+n\cdot 1 = \sum_{j=1}^{n}j(n+1-j)=\dfrac{n(n+1)(n+2)}{6},$$
largest value is
$$
V_n = 1\cdot 1+2\cdot 2+\cdots+n\cdot n = \sum_{j=1}^nj^2= \dfrac{n(n+1)(2n+1)}{6}.
$$ 

Let's prove that there are no "gaps" between $v_n$ and $V_n$.
We'll use math. induction.
0) There are no "gaps" for $n=4$.
Really, 
$(4,3,2,1)\rightarrow 20$;
$(4,2,3,1)\rightarrow 21$; 
$(3,4,1,2)\rightarrow 22$;
$(3,2,4,1)\rightarrow 23$;
$(4,1,2,3)\rightarrow 24$;
$(3,1,4,2)\rightarrow 25$; ... (other cases - symmetrically).
1) Suppose that there are no "gaps" for some $n=n_0$.
2) Now prove that there are no "gaps" for $n=n_0+1$:
2.a) all sums of vectors $(n_0+1, x_1,x_2,\ldots,x_{n_0})$ 
have form
$$
1(n_0+1)+2x_1+3x_2+\cdots+(n_0+1)x_{n_0} \\= n_0+1+(x_1+x_2+\cdots+x_{n_0})+(x_1+2x_2+\cdots+n_0 x_{n_0})
$$
and fill values from 
$$
n_0+1+\dfrac{n_0(n_0+1)}{2}+\dfrac{n_0(n_0+1)(n_0+2)}{6} = \dfrac{(n_0+1)(n_0+2)(n_0+3)}{6}
= v_{n_0+1}$$
to
$$
n_0+1+\dfrac{n_0(n_0+1)}{2}+\dfrac{n_0(n_0+1)(2n_0+1)}{6} = \dfrac{2n_0^2+4n_0+6}{6} 
$$
without "gaps";
2.b) all sums of vectors $(x_1,x_2,\ldots,x_{n_0},n_0+1)$ 
have form
$$
x_1+2x_2+\cdots+n_0x_{n_0}+(n_0+1)(n_0+1) \\= (x_1+2x_2+\cdots+n_0 x_{n_0}) + (n_0+1)^2
$$
and fill values from 
$$
\dfrac{n_0(n_0+1)(n_0+2)}{6} + (n_0+1)^2 = \dfrac{n_0^2+8n_0+6}{6}
$$
to
$$
V_{n_0}+(n_0+1)^2 = \dfrac{(n_0+1)(n_0+2)(2n_0+3)}{6}=V_{n_0+1}.
$$
without "gaps".
2.c) It remains to show that upper bound of 2.a) is not less than lower bound of 2.b). It is so, since 
$$
n_0\ge 4\\ 
\Downarrow \\
n_0^2 \ge 4n_0 \\
\Downarrow \\
2n_0^2 \ge n_0^2+4n_0 \\
\Downarrow \\
2n_0^2 +4n_0+6\ge n_0^2+8n_0+6.
$$

And apply it for $n=100$.