By some erodos result in a graph with $n$ vertices if $|E|\geq \dfrac{(n-1)k}{2}$ then there exist a $\text{K}-{\text{cycle}}$. if $k>2$.So I guess the constraint $k>2$ is missing in the proble as if $k=2$ then for existense of a triangle $\dfrac{n^2}{4}$ edges are needed which is $>$ $\dfrac{n}{2}$ for big $n$ so the result will be false then. Oops my bad

Hi! I passed your remark to the proposer and he says the problem remains true if we allow $k=2$. I believe you read 'cycle' in the problem instead of 'path'.

We will show that every simple graph $G$ with more than $n$ vertices and more than $nk/2$ edges contains a path of length $k+1$. (Here the length of a path is the number of edges it has.) We will use induction on $n\geqslant 2$. So, we can assume $G$ is connected. Assume $k\geqslant 3$. (When $k=2$, $G$ contains a cycle, which must be $C_3$, and an edge connected to the cycle.) For a vertex $v$ of $G$, the graph $G-\{ v\}$ will satisfy the inductive hypothesis iff $e(G)-d(v)>(n-1)k/2$. So, suppose $e(G)-d(v)\leqslant \lfloor (n-1)k/2\rfloor \implies d(v)\geqslant (\lfloor nk/2\rfloor +1)-\lfloor (n-1)k/2\rfloor \geqslant (k+1)/2$. Consider the longest path $v_0,v_1,v_2,\dotsc ,v_t$. It's clear that $t\geqslant 2$. If $v_0v_t\in G$, the vertices create a cycle. If the $t+1$ vertices don't constitute all vertices of $G$, since $G$ is connected, there must be an edge connected to this cycle, creating a longer path. Contradiction. So, $$t+1=n\implies \binom{t+1}{2}>\frac{(t+1)k}{2}\implies t\geqslant k+1.$$ If $v_0v_t\not\in G$, all neighbors of $v_0$ belong to $\{ v_1,v_2,\dotsc ,v_{t-1}\}$. So, there are at least $(k-1)/2$ neighbors of $v_0$ in $\{ v_2,v_3,\dotsc ,v_{t-1}\}$. For each such neighbor $v_r$, $v_{r-1}$ can't be a neighbor of $v_t$. So, $$\frac{k+1}{2}\leqslant d(v_t)\leqslant \left( (t-2)-\frac{k-1}{2}\right) +1\implies t\geqslant k+1.\quad \blacksquare$$ Note that an equivalent formulation of this problem appeared earlier here.