T(n)=n+4n2+8+16n4+32+⋯+Θ(nlg4)=lg(n−1)∑i=0 2in+2lg(n−1)∑i=0 4i+Θ(n2)=2lgn−12−1n+2⋅4lgn−14−1+Θ(n2)=(2lgn−1)n+23(4lgn−1)+Θ(n2)=(n−1)n+23(n2−1)+Θ(n2)=Θ(n2)

Break Down the Problem

1. Identify the expression for $T(n)$.
2. Analyze the individual components of the expression.
3. Combine the terms to simplify $T(n)$.

Relevant Concepts

1. Big Theta notation $\Theta$ indicates asymptotic tight bounds.
2. Summations and logarithmic identities will be used for simplification.
3. The behavior of polynomial and logarithmic terms as $n$ grows.

Analysis and Detail

1. The given expression is: $T(n) = n + 4n^2 + 8 + 16n^4 + 32 + \ldots + \Theta(n \log 4) = \log(n-1) \sum_{i=0}^{2} 2^i n + 2 \log(n-1) \sum_{i=0}^{4} 4^i + \Theta(n^2)$

2. Simplify the summations:

   • The first summation: $\sum_{i=0}^{2} 2^i = 2^0 + 2^1 + 2^2 = 1 + 2 + 4 = 7$ Therefore, $\log(n-1) \sum_{i=0}^{2} 2^i n = 7 \log(n-1) n$

   • The second summation: $\sum_{i=0}^{4} 4^i = 4^0 + 4^1 + 4^2 + 4^3 + 4^4 = 1 + 4 + 16 + 64 + 256 = 341$ Therefore, $2 \log(n-1) \sum_{i=0}^{4} 4^i = 682 \log(n-1)$

3. Combining terms together: $T(n) = 7 \log(n-1) n + 682 \log(n-1) + \Theta(n^2)$

Verify and Summarize

1. Through simplification and collecting the terms, we observe that as $n$ increases:
   • The leading term comes from $\Theta(n^2)$.
   • Thus, $T(n) = \Theta(n^2)$ dominates the logarithmic terms.

Final Answer

$T(n) = \Theta(n^2)$