= Solution
Write $K=k+1$ and $h_j=\mathbb E_jD_j$. The <expected duration of symmetric gambler's ruin> satisfies $h_0=h_K=0$ and, by conditioning on the first step,
$$
h_j=1+\frac12h_{j-1}+\frac12h_{j+1},\qquad1\leq j\leq K-1.
$$
These <expected values> are finite: from any interior state, $K$ consecutive right steps ensure absorption, an event of probability at least $2^{-K}$. Blocks of $K$ steps therefore bound the survival probability by a decreasing geometric sequence. The candidate $h_j=j(K-j)$ has the required boundary values and satisfies $h_{j+1}-2h_j+h_{j-1}=-2$. Hence
$$
\boxed{\mathbb E D_j=j(k+1-j).}
$$
For completeness, the difference of two solutions of this <linear recurrence relation> has zero second difference and is affine in $j$; its two zero boundary values force it to vanish.
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