The Chebyshev polynomial $U_n(x)$ of the second kind is characterized by $U_n(cos\theta)=\frac{\sin(n+1)\theta}{\sin(\theta)}$. It seems that
$$Res_x(U_n(x)+tU_{n-1}(x),\sum_{k=0}^{n-1}U_k(x))=(-1)^{\frac{n(n-1)}{2}}t^{\lfloor\frac{k}{2}\rfloor}2^{n(n-1)},$$
where Res denotes the resultant of two polynomials. 

I do not know how to prove this "equality". Is it a known result? 
Similar results appeared in https://www.researchgate.net/publication/228924014_Resultants_and_discriminants_of_Chebyshev_and_related_polynomials, Theorem 2.

https://www.cambridge.org/core/services/aop-cambridge-core/content/view/D3FA1C0EBDAB83534C86375A0D711BF7/S0008439500017677a.pdf/the-resultant-of-chebyshev-polynomials.pdf