Please help me find discriminant of $\sum_{k=0}^n x^k$. I guess we should find
Res($\sum_{k=0}^n x^k, \sum_{k=1}^n kx^{k-1}$)
by counting an appropriate Sylvester's determinant, but I can't find a simple way to solve it.
This is a task of increased difficulty given on an algebra seminar at HSE Math faculty.
Let $f:=\sum_{k=0}^nX^k$ and note that $(X-1)f=X^{n+1}-1$. Then $$\Delta(X^{n+1}-1)=\Delta(f)\Delta(X-1)\operatorname{Res}(f,X-1)^2.$$ Of course $\Delta(X-1)=1$ and $$\operatorname{Res}(f,X-1)=\operatorname{Res}(f(1),X-1)=\operatorname{Res}(n+1,X-1)=n+1.$$ We can compute $\Delta(X^{n+1}-1)=\operatorname{Res}(X^{n+1}-1,(n+1)X^n)$ as in this question to find that $$\Delta(X^{n+1}-1)=(n+1)^{n+1}(-1)^{(n+1)(n+2)/2+1}.$$ It follows that $$\Delta(f)=(n+1)^{n-1}(-1)^{(n+1)(n+2)/2+1}=(-1)^{(n^2-n)/2}(n+1)^{n-1}.$$
Alternatively, you can compute the discriminant directly. From $(X-1)f=X^{n+1}-1$ we see that the roots of $f$ are precisely $\zeta^k$ for $k=1,\ldots,n$ where $\zeta$ is a primitive $n+1$-th root of unity. Then $$\Delta(f):=\prod_{i<j}(\zeta^i-\zeta^j)^2=(-1)^{n(n+1)/2}\prod_{i\neq j}(\zeta^i-\zeta^j).$$ With the observation that $$\prod_{j=1}^n(1-\zeta^j)=f(1)=n+1,$$ it readily follows, by reindexing the products, that \begin{eqnarray*} \prod_{i\neq j}(\zeta^i-\zeta^j) &=&\prod_{i=1}^n\prod_{\substack{j=1\\ i+j\neq n+1}}^n(\zeta^i-\zeta^{i+j})\\ &=&\prod_{i=1}^n\left((\zeta^i-1)^{-1}\prod_{j=1}^n(\zeta^i-\zeta^{i+j})\right)\\ &=&\prod_{i=1}^n\left(\frac{\zeta^{ni}}{\zeta^i-1}\prod_{j=1}^n(1-\zeta^j)\right)\\ &=&\prod_{i=1}^n\frac{(n+1)\zeta^{ni}}{\zeta^i-1}\\ &=&(n+1)^n\zeta^{n^2(n+1)/2}\frac{1}{\prod_{i=1}^n(1-\zeta^i)}\\ &=&(n+1)^{n-1}\zeta^{n^2(n+1)/2}. \end{eqnarray*} Now note that $\zeta^{n^2(n+1)/2}=1$ if $n$ is even, and otherwise it equals $-1$. So $\zeta^{n^2(n+1)/2}=(-1)^n$ and hence $$\Delta(f)=(-1)^{n(n-3)/2}(n+1)^{n-1}.$$