Constrained Sum - DP Formulation

Let $dp[i]$ = maximum sum of a valid subsequence ending at index $i$ (must include $nums[i]$). The transition: $dp[i] = nums[i] + \max(0, \max_{j=i-k}^{i-1}(dp[j]))$. You take the best previous subsequence sum if it's positive, otherwise start fresh. The $\max(0, ...)$ is what separates this from Jump Game VI.

If all previous subsequences ending in the window have negative sum, you're better off starting a new subsequence at $i$.