1. **State the problem:** We need to solve the recurrence relation $$T(n) = 2T\left(\frac{n}{2} + 17\right) + n$$. 2. **Identify the type of recurrence:** This is a divide-and-conquer recurrence with a non-standard argument inside the recursive call. 3. **Approach:** We can try to analyze the recurrence using the recursion tree method or the Master Theorem, but the addition of 17 inside the argument complicates direct application of the Master Theorem. 4. **Rewrite the recurrence:** Let’s define a new variable to simplify the argument. Set $$m = n + 34$$ so that $$\frac{n}{2} + 17 = \frac{m}{2}$$. 5. **Express T in terms of m:** Then, $$T(n) = 2T\left(\frac{m}{2}\right) + n = 2T\left(\frac{m}{2}\right) + m - 34$$ 6. **Define a new function:** Let $$S(m) = T(m - 34)$$, then $$S(m) = 2S\left(\frac{m}{2}\right) + m - 34$$ 7. **Simplify the recurrence for S(m):** $$S(m) = 2S\left(\frac{m}{2}\right) + m - 34$$ 8. **Apply the Master Theorem:** The recurrence is now $$S(m) = 2S\left(\frac{m}{2}\right) + m - 34$$ which is similar to $$S(m) = 2S\left(\frac{m}{2}\right) + m$$ Here, $a=2$, $b=2$, and $f(m) = m$. 9. **Calculate $\log_b a$:** $$\log_2 2 = 1$$ 10. **Compare $f(m)$ with $m^{\log_b a}$:** $f(m) = m$ and $m^{\log_b a} = m^1 = m$, so $f(m)$ is $\Theta(m^{\log_b a})$. 11. **Master Theorem case 2 applies:** When $f(m) = \Theta(m^{\log_b a} \log^k m)$ with $k=0$, the solution is $$S(m) = \Theta(m^{\log_b a} \log^{k+1} m) = \Theta(m \log m)$$ 12. **Translate back to $T(n)$:** Since $m = n + 34$, asymptotically $m \sim n$, so $$T(n) = S(n + 34) = \Theta(n \log n)$$ **Final answer:** $$\boxed{T(n) = \Theta(n \log n)}$$