# $f:[a,b]\to(a,b)$ be continuous how prove $f(c)+f(c+d)+\cdots+f(c+nd)=(n+1)(c+\frac{nd}{2})$

let $f:[a,b]\to(a,b)$ be continuous how prove $\forall n\in\mathbb N$ $\exists d\gt0$ ,$\exists c\in(a,b)$ such that $$f(c)+f(c+d)+\cdots+f(c+nd)=(n+1)\left(c+\frac{nd}{2}\right)$$thanks in advance

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Let $n \in \mathbb{N}$ be fixed and write $g(x,y) = \sum_{k=0}^n f(x + ky) - (n+1)(x + ny/2)$. Then $g$ is a continuous function $\Delta \to \mathbb{R}$, where $\Delta$ is the triangle given by the inequalities $$a \le x \le b, \quad 0 \le y \le \frac{b-x}{n}.$$ Notice that since $f \colon [a,b] \to (a,b)$, we have $$g(a,0) = (n+1)f(a) - (n+1)a > 0,$$ $$g(b,0) = (n+1)f(b) - (n+1)b < 0.$$ By continuity there exist neighbourhoods $U$ of $(a,0)$ and $V$ of $(b,0)$ in $\Delta$ such that $g(x,y) > 0$ for all $(x,y) \in U$ and $g(x,y) < 0$ for all $(x,y) \in V$. In particular there exist $d > 0$ and $a < b' < b$ such that $(a,d) \in U$ and $(b',d) \in V$. Now by Bolzano's theorem there exists $c \in (a,b')$ such that $g(c,d) = 0$ and the claim follows.