Answer: \(\frac{1}{3 \times 4^{n-1}}\).
Step-by-step explanation:
To simplify the expression \(4^{-2n} \times 2^{2n+1} \times 3^n \div 3^{n+1} \times 2^{-2n+1}\), let's break it down step by step:
1. Start with the powers of 4 and 2:
\(4^{-2n} = \frac{1}{4^{2n}} = \frac{1}{(2^2)^{2n}} = \frac{1}{2^{4n}}\)
\(2^{2n+1}\) stays the same.
\(2^{-2n+1} = \frac{1}{2^{2n-1}}\)
2. Combine the terms:
\(\frac{1}{2^{4n}} \times 2^{2n+1} \times 3^n \div 3^{n+1} \times \frac{1}{2^{2n-1}}\)
3. Combine the terms with the same base:
\(2^{2n+1 - (4n - 1)} \times 3^{n - (n+1)}\)
\(= 2^{2n+1 - 4n + 1} \times 3^{n - n - 1}\)
\(= 2^{-2n+2} \times 3^{-1}\)
4. Simplify further:
\(= \frac{1}{2^{2n-2}} \times \frac{1}{3}\)
\(= \frac{1}{4^{n-1}} \times \frac{1}{3}\)
\(= \frac{1}{3 \times 4^{n-1}}\)
So, the simplified expression is \(\frac{1}{3 \times 4^{n-1}}\).